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The ''ratio test'' allows for proving convergence or divergence of many explicitly given series, so it is among the most popular criteria in use.
Although it is applicable to fewer series than the root test, proofs based on the ratio test are usually easier to do.

The ratio test was first published by mathematician and physicist [[w:en:Jean_le_Rond_d'Alembert|Jean-Baptiste le Rond d’Alembert]] and is thus sometimes called ''d'Alembert's ratio test''.

== Derivation ==
=== First thoughts ===
Similar to the root test, the ratio test makes use of the direct comparison test to reduce the convergence of the series in question to that of a geometric series.
Let <math>\sum_{k=1}^\infty a_k</math> be a given series with <math>a_k\ge 0</math> for all <math>k\in\N</math>.
The requirement of non-negative summands is necessary for the direct comparison test.
We know that the series converges, if there is some <math>q\in[0,1)</math> such that <math>a_k\le q^k</math> for all <math>k\in\N</math>.
This follows immediately from direct comparison to the geometric series <math>\sum_{k=1}^\infty q^k</math>, which converges for <math>0\le q < 1</math>.

The root test simply transforms the inequality <math>a_k\le q^k</math> to <math>\sqrt[k]{a_k} \le q</math>.
The ratio test instead employs a recursive argument with <math>a_k\le q^k</math> as an implication.
As a starting point, we require <math>a_1 \le q</math> (so the inequality holds for the base case <math>k=1</math>).
To prove the target inequality for all <math>k</math> by induction, we would need a criterion allowing to deduce <math>a_{k+1} \le q^{k+1}</math> from the inductive assumption <math>a_k \le q^k</math>.
Assuming <math>a_k\neq 0</math>, we find

{{Math|<math>a_{k+1} = \frac{a_{k+1}}{a_k}\cdot a_k \le \frac{a_{k+1}}{a_k}\cdot q^k,</math>}}

where we used that <math>\tfrac{a_{k+1}}{a_k}</math>, as a ratio of non-negative numbers, is itself non-negative.
Since we already assumed <math>a_k\ge 0</math>, the set of series for which the ratio test is applicable reduces to those with <math>a_k > 0</math> for all <math>k\in\N</math>.

As a consequence, to deduce <math>a_{k+1} \le q^{k+1}</math> from the inductive assumption it suffices to have <math>\tfrac{a_{k+1}}{a_k}\cdot q^k \le q^{k+1}</math>.
In turn, a sufficient condition for this to hold is the simple recursive relation

{{Math|<math>0 < \frac{a_{k+1}}{a_k} \le q.</math>}}

=== Summarizing our first thoughts ===
Assuming <math>a_k > 0</math>, we can show inductively that <math>a_1 \le q</math> and <math>\tfrac{a_{k+1}}{a_k} \le q</math> together imply <math>a_k \le q^k</math> for all <math>k\in\N</math>.
This statement is a direct comparison with a geometric series. Such series converge for <math>q \in [0,1)</math>, and so does the series in question, if all criteria are met.
Given the base case <math>a_1 \le q</math> and inductive assumption <math>a_k \le q^k</math>, the inductive step reads:

{{Math|<math>\begin{align}
a_{k+1} & = \frac{a_{k+1}}{a_k}\cdot a_k \\[0.5em]
&\ {\color{OliveGreen}\left\downarrow\ a_k \le q^k \right.} \\[0.5em]
& \le \frac{a_{k+1}}{a_k}\cdot q^k \\[0.5em]
&\ {\color{OliveGreen}\left\downarrow\ \frac{a_{k+1}}{a_k} \le q \right.} \\[0.5em]
& \le q^{k+1} \\[0.5em]
\end{align}</math>}}

=== First improvement ===
Whether a series converges or diverges does not depend on finitely many of its summands, as convergence is a property of the infinite.
That means, if we take a convergent series and change a finite number of its summands, we obtain another convergent series (though with a possibly different value).
Hence one could expect the requirement <math>a_1 \le q</math> to be irrelevant for the convergence of the entire series, as it only affects a single summand.

In fact, assuming only <math>\tfrac{a_{k+1}}{a_k} \le q</math> we find

{{Math|<math>\begin{align}
a_1 &&&&= a_1 \cdot q^0, \\
a_2 &= a_1 \cdot \frac{a_2}{a_1} &&\le a_1 \cdot q &= a_1 \cdot q^1, \\
a_3 &= a_2 \cdot \frac{a_3}{a_2} &&\le a_2 \cdot q &\le a_1 \cdot q^2, \\
a_4 &= a_3 \cdot \frac{a_4}{a_3} &&\le a_3 \cdot q &\le a_1 \cdot q^3, \\
a_5 &= a_4 \cdot \frac{a_5}{a_4} &&\le a_4 \cdot q &\le a_1 \cdot q^4, \\
&&\vdots
\end{align}</math>}}

Altogether, we have <math>a_k \le a_1 \cdot q^{k-1}</math>.
The series <math>\sum_{k=1}^\infty a_1 \cdot q^{k-1} = \tfrac{a_1}{q} \cdot \sum_{k=1}^\infty q^k</math> is proportional to a convergent geometric series, so it is itself convergent.
By direct comparison, we have thus shown convergence of the series <math>\sum_{k=1}^\infty a_k</math> using <math>\tfrac{a_{k+1}}{a_k} \le q</math> alone, without any restrictions to <math>a_1</math>.

=== Second improvement ===
We can generalize our induction proof even further by not only dropping our special requirement for <math>a_1</math>, but also the quotient requirement for a ''finite'' number of pairs of subsequent summands.
In mathematical terms: We require <math>\tfrac{a_{k+1}}{a_k}\le q</math> only for ''almost all'' <math>k\in\N</math>.
After that, we are still left with an infinite number of pairs for which the requirement holds.
In particular, we can find some <math>K\in\N</math>, such that the criterion still holds for ''all'' <math>k\ge K</math>.
Beyond this index we have a similar situation as before:

{{Math|<math>\begin{align}
a_K &&&&= a_K \cdot q^0 \\[0.5em]
a_{K+1} &= a_K     \cdot \frac{a_{K+1}}{a_K}     &\le a_K     \cdot q &&=   a_K \cdot q^1 \\[0.5em]
a_{K+2} &= a_{K+1} \cdot \frac{a_{K+2}}{a_{K+1}} &\le a_{K+1} \cdot q &&\le a_K \cdot q^2 \\[0.5em]
a_{K+3} &= a_{K+2} \cdot \frac{a_{K+3}}{a_{K+2}} &\le a_{K+2} \cdot q &&\le a_K \cdot q^3 \\[0.5em]
a_{K+4} &= a_{K+3} \cdot \frac{a_{K+4}}{a_{K+3}} &\le a_{K+3} \cdot q &&\le a_K \cdot q^4 \\[0.5em]
&\vdots
\end{align}</math>}}

Altogether we have <math>a_{K+l} \le a_K \cdot q^l</math> for all <math>l \ge 0</math>.
After an index shift <math>k=K+l</math> the inequality reads <math>a_k \le a_K\cdot q^{k-K}</math> for all <math>k \ge K</math>.
We can now find a finite upper estimate for the whole series:

{{Math|<math>\begin{align}
\sum_{k=1}^\infty a_k &= \sum_{k=1}^{K-1} a_k + \sum_{k=K}^\infty a_k \\[0.5em]
&{\color{OliveGreen}\left\downarrow\ \forall k \ge K: a_k \le a_K \cdot q^{k-K} \right.} \\[0.5em]
&\le \sum_{k=1}^{K-1} a_k + \sum_{k=K}^\infty a_K \cdot q^{k-K} \\[0.5em]
&= \underbrace{\underbrace{\sum_{k=1}^{K-1} a_k}_{\text{finite}} + a_K \underbrace{\sum_{k=0}^\infty q^k}_{\text{convergent}}}_{\text{finite}} \\[0.5em]
\end{align}</math>}}

This proves convergence of the series by direct comparison. Hence, it is successful to require <math>\tfrac{a_{k+1}}{a_k} \le q</math> only for almost all <math>k\in\N</math>.

=== Rephrasing in terms of limit superior ===
The condition <math>\tfrac{a_{k+1}}{a_k} \le q</math> for almost all <math>k\in\N</math> and some fixed <math>q</math> with <math>0<q<1</math> can be expressed using the notion of limit superior. In fact, the statement of the previous phrase is equivalent to <math>\limsup_{k\to\infty} \tfrac{a_{k+1}}{a_k} < 1</math>.

We prove equivalence by starting with the first statement. <math>\tfrac{a_{k+1}}{a_k} \le q</math> for almost all <math>k</math> implies that all cluster points of the sequence <math>\left(\tfrac{a_{k+1}}{a_k}\right)_{k\in\N}</math> must be smaller than or equal to <math>q</math>. In particular, the limit superior, which is the greatest cluster point, then must obey <math>\limsup_{k\to\infty} \tfrac{a_{k+1}}{a_k} \le q < 1</math>, which is the second statement.

Now for the converse direction. Let <math>\tilde{q} \equiv \limsup_{k\to\infty} \tfrac{a_{k+1}}{a_k} < 1</math>. Then for any <math>\epsilon > 0</math> the inequality <math>\tfrac {a_{k+1}}{a_k} \le \tilde q + \epsilon</math> holds for almost all <math>k\in\N</math>. Since <math>0 < \tilde{q} < 1</math>, we can choose <math>\epsilon > 0</math> small enough such that also <math>0 < \tilde{q} + \epsilon < 1</math>, e.g. <math>\epsilon \equiv \tfrac{1-\tilde{q}}{2}</math>. If we now set <math>q \equiv \tilde{q} + \epsilon</math>, we have both <math>0 < q < 1</math> and <math>\tfrac{a_{k+1}}{a_k} \le q</math> for almost all <math>k\in\N</math>. Given the second statement, we can thus explicitly construct a <math>q</math> for which the first statement holds.

We can summarize that <math>\limsup_{k\to\infty} \tfrac{a_{k+1}}{a_k} < 1</math> is a sufficient criterion for the series <math>\sum_{k=1}^\infty a_k</math> to converge.

=== Adding a flavor of absolute convergence ===
So far, we restricted ourselves to series with non-negative summands. Can we extend our convergence criterion to general series with (at least some) negative summands?

For any given series <math>\sum_{k=1}^\infty a_k</math>, we can construct another series <math>\sum_{k=1}^\infty |a_k|</math>, whose summands are clearly non-negative. This series is now in the range of applicability for our ratio test. However, showing convergence for that series is exactly what it means to show ''absolute convergence'' for the original series. As we have seen before, absolute convergence implies "common" convergence.

{{Important|If <math>\limsup_{k\to\infty} \tfrac{|a_{k+1}|}{|a_k|} = \limsup_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| < 1</math>, then the series <math>\sum_{k=1}^\infty a_k</math> is absolutely convergent, and hence convergent.}}

Introduction of the absolute value changes nothing for series whose summands have been non-negative already (the situation we assumed so far). Thus, the new version of the ratio test introduced in this section is strictly more powerful than the one we considered before, as it has a larger range of applicability and absolute convergence is a stronger statement than convergence.

=== Ratio test for divergence ===
Is it possible to prove the divergence of a series with a similar argument? Let's look at <math>\left|\tfrac{a_{k+1}}{a_k}\right|</math>. If the (absolute value of the) quotient is greater than or equal to one, then

{{Math|<math> \left|\frac{a_{k+1}}{a_k}\right| \ge 1 \implies  |a_{k+1}| \ge |a_k|</math>}}

Thus, if starting from any index <math>K</math> for all subsequent indices <math>k</math> the inequality <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> is satisfied, then the sequence <math>\left(|a_k|\right)_{k\in\N}</math> grows monotonically, starting from the index <math>K</math>. This sequence cannot be a zero sequence, because it grows monotonically after the sequence member <math>|a_K|</math> and <math>|a_K|> 0</math>. But if <math>\left(|a_k|\right)_{k\in\N}</math> is not a zero sequence, then <math>\left(a_k\right)_{k\in\N}</math> is not a zero sequence either. It follows, according to the term test, that the series <math>\sum_{k=1}^\infty a_k</math> is not a zero sequence. After all, the term test states that <math>\lim_{k\to\infty} a_k= 0</math> would hold if the series <math>\sum_{k=1}^\infty a_k</math> was convergent. To summarise:

{{Important|If <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> holds for almost all <math>k\in\N</math> , then <math>\left(a_k\right)_{k\in\N}</math> is not a null sequence. The series <math>\sum_{k=1}^\infty a_k</math> diverges by the term test.}}

== The ratio test ==
=== Theorem ===
[[File:Quotienten-Kriterium - Quatematik.webm|thumb|Explanation of the ratio test. ([https://www.youtube.com/watch?v=EkJ8OLidE14 YouTube video] (in German) by the channel [https://www.youtube.com/channel/UC0LfPSALkn8YDhXZf8-Pl5A Quatematik])]]
{{:Math for Non-Geeks: Template:Satz
 |titel=Ratio test for divergence
 |satz=
Let <math>\sum_{k=1}^\infty a_k</math> be a series with <math>a_k\neq0</math> for all <math>k\in\N</math>. If <math>\limsup_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| < 1</math>, then the series <math>\sum_{k=1}^\infty a_k</math> is absolutely convergent. 

If <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> holds for almost all <math>k\in\N</math> (that is, for all <math>k\ge K</math> for a given <math>K\in\N</math>), then the series <math>\sum_{k=1}^\infty a_k</math> is divergent.

  |erklärung=
We summarise the above derivation in a proof:

  |beweis=
Let <math>\sum_{k=1}^\infty a_k</math> be a series with non-zero summands.

{{:Math for Non-Geeks: Template:Beweisschritt
 |ziel=Ratio test for convergence
 |beweisschritt=
Let <math> \theta = \limsup_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| < 1</math>. We now choose <math> \epsilon>0</math> so small that <math>q= \theta + \epsilon < 1</math>. Because of <math> \theta < 1</math> this <math>\epsilon</math> exists (for example, <math>\epsilon = \tfrac{1- \theta}2</math> can be chosen). From the properties of the limit superior it follows that for almost all <math>k\in\N</math> the inequality <math>\left|\tfrac{a_{k+1}}{a_k}\right| < q</math> is satisfied. So there is a natural number <math>K</math> such that <math>\left|\tfrac{a_{k+1}}{a_k}\right| < q</math> for all <math>k\ge K</math>. Hence:

{{Math|<math>\begin{align}
|a_{K+1}| & = |a_K|     \cdot \left|\frac{a_{K+1}}{a_K}\right|     \le |a_K|     \cdot q \\[0.5em]
|a_{K+2}| & = |a_{K+1}| \cdot \left|\frac{a_{K+2}}{a_{K+1}}\right| \le |a_{K+1}| \cdot q  \le |a_K| \cdot q^2 \\[0.5em]
|a_{K+3}| & = |a_{K+2}| \cdot \left|\frac{a_{K+3}}{a_{K+2}}\right| \le |a_{K+2}| \cdot q  \le |a_K| \cdot q^3 \\[0.5em]
|a_{K+4}| & = |a_{K+3}| \cdot \left|\frac{a_{K+4}}{a_{K+3}}\right| \le |a_{K+3}| \cdot q  \le |a_K| \cdot q^4 \\[0.5em]
& \vdots
\end{align}</math>}}

Altogether, we obtain <math>|a_{K+l}| \le |a_K| \cdot q^l</math>. By setting <math>k=K+l</math>, the inequality <math>|a_k| \le |a_K| \cdot q^{k-K}</math> follows and thus:

{{Math|<math>\begin{align}
\sum_{k=1}^\infty |a_k| & = \sum_{k=1}^{K-1} |a_k| + \sum_{k=K}^\infty |a_k| \\[0.5em]
& {\color{OliveGreen}\left\downarrow\ \forall k \ge K: |a_k| \le |a_K| \cdot q^{k-K} \right.} \\[0.5em]
&\le \sum_{k=1}^{K-1} |a_k| + \sum_{k=K}^\infty |a_K| \cdot q^{k-K} \\[0.5em]
& {\color{OliveGreen}\left\downarrow\ \text{index shift in the 2nd series } \right.} \\[0.5em]
&= \underbrace{\underbrace{\sum_{k=1}^{K-1} |a_k|}_{\text{bounded}} + \underbrace{|a_K| \sum_{k=0}^\infty q^k}_{\text{convergent = bounded}}}_{\text{bounded}} \\[0.5em]
\end{align}</math>}}

So the series <math>\sum_{k=1}^\infty |a_k|</math> converges according to the direct comparison test. This in turn means that <math>\sum_{k=1}^\infty a_k</math> converges absolutely.
}}

{{:Math for Non-Geeks: Template:Beweisschritt
 |ziel=Ratio test for divergence
 |beweisschritt=
Now let <math>K</math> be a natural number such that <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> for almost all <math>k\ge K</math>. Then for all <math>k\ge K</math> there is:

{{Math|<math>\begin{align}
&& \left|\frac{a_{k+1}}{a_k}\right| & \ge 1 \\[0.5em]
& \implies & |a_{k+1}| & \ge |a_k|
\end{align}
</math>}}

Thus the sequence <math>\left(|a_k|\right)_{k\in\N}</math> grows monotonically starting from the index <math>K</math>. The sequence <math>\left(|a_k|\right)_{k\in\N}</math> is not zero because <math>|a_K| > 0</math> (as <math>a_K \neq 0</math>). But this means that <math>\left(a_k\right)_{k\in\N}</math> is also not a zero sequence. From the term test it follows that the series <math>\sum_{k=1}^\infty a_k</math> diverges.
}}}}

=== Stronger statement by the limes inferior ===
The condition for divergence which we just discussed can be tightened using the limes inferior. This makes the criterion easier to apply. If <math>\liminf_{k\to \infty} \left| \tfrac{a_{k+1}}{a_k}\right|>1</math>, it follows that <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> for almost all <math>k\ge K</math>. So the series diverges. The converse does not always hold true. From <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> for almost all <math>k\ge K</math> we cannot imply <math>\liminf_{k\to \infty} \left| \tfrac{a_{k+1}}{a_k}\right|>1</math>, since the sequence <math>\left(\left|\tfrac{a_{k+1}}{a_k}\right|\right)_{k\in\N}</math> does not necessarily have a smallest accumulation point. It is therefore a stronger condition for the divergence of the series.

{{:Math for Non-Geeks: Template:Hinweis|If <math>\lim_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| < 1</math>, then there exists a <math>\theta < 1</math> with <math>\left|\tfrac{a_{k+1}}{a_k}\right| < \theta</math> for almost all <math>k\in\N</math>, so the series converges absolutely. Similarly, the series diverges if <math>\lim_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| > 1</math>.}}

== Limits of the ratio test ==
For <math>\lim_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| = 1</math> we cannot say anything about convergence or divergence of the series. There are in fact both convergent and divergent series that fulfil this condition. An example of this is the divergent series <math>\sum_{k=1}^\infty \frac 1k</math>:

{{Math|<math>\begin{align}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| & = \lim_{k\to\infty} \frac{k}{k+1} \\[0.5em]
& = \lim_{k\to\infty} \frac{1}{1+\frac 1k} = 1
\end{align}</math>}}

The convergent series <math>\sum_{k=1}^\infty \frac 1{k^2}</math> also satisfies this equation:

{{Math|<math>\begin{align}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| & = \lim_{k\to\infty} \frac{k^2}{(k+1)^2} \\[0.5em]
& = \lim_{k\to\infty} \frac{k^2}{k^2+2k+1} \\[0.5em]
& = \lim_{k\to\infty} \frac{1}{1+\frac 2k+\frac 1{k^2}} = 1
\end{align}</math>}}

So from <math>\lim_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| = 1</math> we can neither conclude that the series converges nor that it diverges. We have to use a different convergence criterion in such a case.

== Conducting the ratio test ==
[[File:Decision_diagram_for_the_ratio_test.svg|thumb|class=noprint|Decision tree for the ratio test]]
In order to apply the ratio test to a series <math>\sum_{k=1}^\infty a_k</math>, we first form <math>\left|\tfrac{a_{k+1}}{a_k}\right|</math> and consider the limit:

# If <math>\limsup_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| < 1</math>, then the series converges absolutely.
# If <math>\liminf_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| > 1</math>, then the series diverges.
# If <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> for almost all <math>k\in\N</math>, then the series diverges.
# If we cannot apply any of the three cases, we cannot say anything about the convergence of the series.

== Exercises ==
=== Exercise 1 ===
{{:Math for Non-Geeks: Template:Aufgabe
 |aufgabe=
Investigate whether the series <math>\sum_{k=1}^\infty \frac{2^k}{k!}</math> converges or diverges.

 |lösungsweg=
First we form the quotient <math>\left|\tfrac{a_{k+1}}{a_k}\right|</math> and consider its limit:

{{Math|<math>\begin{align}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| & = \lim_{k\to\infty} \frac{2^{k+1}}{(k+1)!} \cdot \frac{k!}{2^k} \\[0.5em]
&= \lim_{k\to\infty} \frac{2}{k+1} \\[0.5em]
&= 0
\end{align}</math>}}

Thus <math>\lim_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| < 1</math>, with which it follows from the quotient criterion that the series converges absolutely.

 |beweis=
The series <math>\sum_{k=1}^\infty \frac{2^k}{k!}</math> converges absolutely according to the ratio test, as

{{Math|<math>\begin{align}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| & = \lim_{k\to\infty} \frac{2^{k+1}}{(k+1)!} \cdot \frac{k!}{2^k} \\[0.5em]
&= \lim_{k\to\infty} \frac{2}{k+1} \\[0.5em]
&= 0 < 1
\end{align}</math>}}
}}

=== Exercise 2 ===
{{:Math for Non-Geeks: Template:Aufgabe
 |aufgabe=
Investigate whether the series <math>\sum_{k=1}^\infty \frac{k^k}{k!}</math> converges or diverges.

 |lösungsweg=
There is <math>a_k=\tfrac{k^k}{k!}</math>. Now, let us take a look at the quotient <math>\left|\tfrac{a_{k+1}}{a_k}\right|</math>

{{Math|<math>\begin{align}
\left|\frac{a_{k+1}}{a_k}\right| & = \left|\frac{(k+1)^{k+1}}{(k+1)!} \cdot \frac{k!}{k^k}\right| \\[0.5em]
&\ {\color{OliveGreen} \left\downarrow\ \frac{k!}{(k+1)!} = \frac{1}{k+1} \right.} \\[0.5em]
&= \frac{(k+1)^{k+1}}{(k+1)\cdot k^k} \\[0.5em]
&= \frac{(k+1)^k}{k^k} \\[0.5em]
&= \left(1+\frac 1k\right)^k
\end{align}</math>}}

Now <math>1+\tfrac 1k\ge 1</math> and therefore also <math>\left(1+\tfrac 1k\right)^k \ge 1</math>. It follows that <math>\left|\tfrac{a_{k+1}}{a_k}\right| \ge 1</math> for all <math>k\in\N</math> and thus in particular for almost all <math>k\in\N</math>. The ratio test now yields that the series diverges.

 |beweis=
The series <math>\sum_{k=1}^\infty \frac{k^k}{k!}</math> diverges, because for <math>a_k=\tfrac{k^k}{k!}</math> we have

{{Math|<math>\begin{align}
\left|\frac{a_{k+1}}{a_k}\right| & = \left|\frac{(k+1)^{k+1}}{(k+1)!} \cdot \frac{k!}{k^k}\right| \\[0.5em]
&\ {\color{OliveGreen} \left\downarrow\ \frac{k!}{(k+1)!} = \frac{1}{k+1} \right.} \\[0.5em]
&= \frac{(k+1)^{k+1}}{(k+1)\cdot k^k} \\[0.5em]
&= \frac{(k+1)^k}{k^k} \\[0.5em]
&= \left(1+\frac 1k\right)^k \ge 1
\end{align}</math>}}
}}

{{:Math for Non-Geeks: Template:Hinweis|You may already know that <math>\lim_{k\to\infty} \left(1+\tfrac 1k\right)^k = e</math>. Accordingly, you can alternatively prove that <math>\lim_{k\to\infty} \left|\tfrac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \left(1+\tfrac 1k\right)^k = e > 1</math>. However, this reasoning can mathematically only be applied after <math>\lim_{k\to\infty} \left(1+\tfrac 1k\right)^k = e</math> has been proven within an analysis course.}}

=== Exercise 3 ===
{{:Math for Non-Geeks: Template:Aufgabe
 |aufgabe=
Investigate for which <math>a\in\R</math> the series <math display="inline">\sum_{k=0}^\infty \frac{(a^2-1)^k}{k+1}</math> converges (absolutely) or diverges. 

 |beweis=
We use the ratio test with <math display="inline">a_k=\frac{(a^2-1)^k}{k+1}</math>:

{{Math|<math>\begin{align}
\left | \frac{a_{k+1}}{a_k}\right | =&\left| \frac{(a^2-1)^{k+1}\cdot (k+1)}{(a^2-1)^k\cdot (k+2)} \right| \\
&= \left|a^2-1\right|\cdot \frac{k+1}{k+2} \\
&= \left|a^2-1\right|\cdot \frac{1+\frac 1 k}{1+\frac 2 k}
\end{align}</math>}}

Thus

{{Math|<math>\begin{align}
\lim_{k\to\infty}\left| \frac{a_{k+1}}{a_k}\right |= \lim_{k\to\infty} \left|a^2-1\right|\cdot \frac{1+\frac 1 k}{1+\frac 2 k}=\left|a^2-1\right|
\end{align}</math>}}

Because the limit exists, limes superior and limes inferior coincide. Therefore

{{Math|<math>\begin{align}
\limsup_{k\to\infty} \left | \frac{a_{k+1}}{a_k}\right | =\liminf_{k\to\infty} \left | \frac{a_{k+1}}{a_k}\right | =\lim_{k\to\infty}\left | \frac{a_{k+1}}{a_k}\right|=\left|a^2-1\right|
\end{align}</math>}}

We are interested in for which <math>a\in\R</math> there is convergence, absolute convergence and divergence. From the ratio test it follows that the series converges absolutely if <math>|a^2-1|<1</math>. In the case <math>|a^2-1|>1</math> the ratio test yields that the series diverges. The case <math>|a^2-1|=1</math> must be examined separately: 

{{:Math for Non-Geeks: Template:Fallunterscheidung
 |fall1=<math>|a^2-1|<1</math>
 |beweis1=
We want to find out for which <math>a\in\R</math> the inequality <math>|a^2-1|<1</math> holds. By transformations we find:

{{Math|<math>\begin{array}{rrclcrcl}
     &      & |a^2-1| & < 1 \\
\iff & -1 < & a^2-1   & < 1 \\
\iff & 0  < & a^2     & < 2 \\
\iff & 0  < & |a|     & < \sqrt{2} \\
\iff & -\sqrt 2  < & a & < 0 & \text{ or } & 0 < & a & <\sqrt 2 \\
\end{array}</math>}}

For <math>-\sqrt 2<a<0</math> or <math>0< a < \sqrt 2</math>, so if <math>a\in (-\sqrt 2,0) \cup (0,\sqrt 2)</math> holds, then the series converges absolutely.

 |fall2=<math>|a^2-1|>1</math>
 |beweis2=Here, we have:
 
{{Math|<math>\begin{array}{rrlcrl}
      & |a^2-1| & > 1 \\
\iff & a^2-1 & > 1  & \text{ or } & a^2-1 & < -1 \\
\iff & a^2 & > 2  & \text{ or } & a^2 & < 0 \\
\iff & |a| & > \sqrt 2  \\
\iff & a & > \sqrt 2  & \text{ or } & a & < -\sqrt 2
\end{array}</math>}}

For <math>a<-\sqrt 2</math> or <math>a > \sqrt 2</math> , i.e., for <math>a\in (-\infty,-\sqrt 2) \cup (\sqrt 2,\infty)</math> the series diverges

 |fall3=<math>|a^2-1|=1</math>
 |beweis3=And we have that

{{Math|<math>\begin{array}{rrlcrl}
      & |a^2-1| & = 1 \\
\iff & a^2-1 & = 1  & \text{ or } & a^2-1 & = -1 \\
\iff & a^2 & = 2  & \text{ or } & a^2 & = 0 \\
\iff & |a| & = \sqrt 2 & \text{ or } & |a| & = 0 \\
\iff & a & = \pm \sqrt 2  & \text{ or } & a & = 0 
\end{array}</math>}}

Since the ratio test does not provide a convergence statement here, we have to examine the two cases individually:

{{:Math for Non-Geeks: Template:Fallunterscheidung
 |fall1=<math>a=\pm \sqrt 2</math>
 |beweis1=
We have

{{Math|<math>\sum_{k=0}^\infty \frac{(a^2-1)^k}{k+1} = \sum_{k=0}^\infty \frac{1^k}{k+1} = \sum_{k=0}^\infty \frac{1}{k+1}</math>}}

Since this is a harmonic series, it diverges.

 |fall2=<math>a=0</math>
 |beweis2=
We have

{{Math|<math>\sum_{k=0}^\infty \frac{(a^2-1)^k}{k+1} = \sum_{k=0}^\infty \frac{(-1)^k}{k+1}</math>}}

The series is convergent according to the Leibniz criterion, but not absolutely convergent, since <math>\sum_{k=0}^\infty \left|\frac{ (-1)^k}{k+1}\right| = \sum_{k=0}^\infty \frac{1}{k+1}</math> diverges as it is a harmonic series.
}}}}}}

== Comparison: ratio and root test ==
The ratio test is much easier to apply to some series than the root test. An example is the series <math display="inline">\sum_{k=1}^\infty \frac{2^k}{k!}</math>, whose convergence can be well proven with the ratio test:

{{Math|<math>\begin{align}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \frac{2^{k+1}}{(k+1)!} \cdot \frac{k!}{2^k} = \lim_{k\to\infty} \frac{2}{k+1} = 0
\end{align}</math>}}

For the root test, we must consider the following limit:

{{Math|<math>\lim_{k\to\infty} \sqrt[k]{|a_k|} = \lim_{k\to\infty} \frac{2}{\sqrt[k]{k!}}</math>}}

Here it is unclear whether there is convergence and against what. The fact that <math>k!</math> grows rapidly could speak for a zero sequence. However, the sequence <math>\left(\sqrt[k]{k!}\right)</math> is "weakened" by taking the <math>k</math>-th root. In fact, <math>\sqrt[k]{k!} \to \infty</math> can be shown (and thus <math>\sqrt[k]{|a_k|} \to 0</math> follows). However, this proof is very laborious. The situation is similar with the series <math display="inline">\sum_{k=1}^\infty \frac{k^k}{k!}</math>. By the ratio test,

{{Math|<math>\begin{align}
\left|\frac{a_{k+1}}{a_k}\right| & = \left|\frac{(k+1)^{k+1}}{(k+1)!} \cdot \frac{k!}{k^k}\right| \\[0.5em]
&\ {\color{OliveGreen} \left\downarrow\ \frac{k!}{(k+1)!} = \frac{1}{k+1} \right.} \\[0.5em]
&= \frac{(k+1)^{k+1}}{(k+1)\cdot k^k} = \frac{(k+1)^k}{k^k} \\[0.5em]
&= \left(1+\frac 1k\right)^k \ge 1
\end{align}</math>}}

Thus the sequence is divergent according to the ratio test.

For the root test we have to consider the following limit value:

{{Math|<math>\lim_{k\to\infty} \sqrt[k]{|a_k|} = \lim_{k\to\infty} \frac{k}{\sqrt[k]{k!}}</math>}}

One can prove that this sequence converges to <math>e</math>. However, this is laborious and requires additional convergence criteria that are often not available in a basic real analysis lecture. In both cases, the solution with the ratio test is easier.

However, there are also series that can be successfully investigated with the root criterion and for which the ratio test is not applicable. An example of this is the series

{{Math|<math>\sum_{k=0}^\infty a_k \text{ with } a_k = \begin{cases} \left( \frac 12 \right)^k & \text{ for even } , \\ 
\left( \frac 13 \right)^k & \text{ for odd } k\end{cases}</math>}}

The ratio test is not applicable here: For the quotient sequence,

{{Math|<math>\left| \frac{a_{k+1}}{a_k} \right| = \begin{cases} \left( \frac 23 \right)^k & \text{ for even } k \\ 
\left( \frac 32 \right)^k & \text{ for odd } k\end{cases}</math>}}

Thus <math display="inline">\limsup_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \infty</math>, since the quotient sequence for odd <math>k</math> is unbounded from above as <math>\tfrac 32 > 1</math>. On the other hand, <math display="inline">\left| \frac{a_{k+1}}{a_k} \right|<1</math> for all even <math>k</math> and thus for infinitely many quotients. Overall, however, we have to conclude that the ratio test is not applicable. On the other hand, the root test yields

{{Math|<math>\sqrt[k]{|a_k|} = \begin{cases} \frac 12  & \text{ for even } k \\ 
\frac 13 & \text{ for odd } k\end{cases}</math>}}

Thus <math display="inline">\limsup_{k\to \infty} \sqrt[k]{|a_k|} = \frac 12 <1</math> and the series converges absolutely. So in the above example, the root test is applicable, while the ratio test gives no reasonable result.

In general, the root test has even a wider range of application than the ratio test: The root test can be applied to every series where the ratio test is successful. This is a consequence of the following inequality:

{{Math|<math>\liminf_{k\to \infty} \left| \frac{a_{k+1}}{a_k} \right| \leq \liminf_{k\to \infty} \sqrt[k]{|a_k|} \leq \limsup_{k\to \infty} \sqrt[k]{|a_k|} \leq \limsup_{k\to \infty} \left| \frac{a_{k+1}}{a_k} \right|</math>}} 

Here it becomes obvious: If <math display="inline">\limsup_{k\to \infty} \left| \frac{a_{k+1}}{a_k} \right|<1</math>, then automatically <math display="inline">\limsup_{k\to \infty} \sqrt[k]{|a_k|}<1</math>. If <math display="inline">\liminf_{k\to \infty} \left| \frac{a_{k+1}}{a_k} \right|>1</math>, then automatically <math display="inline">\limsup_{k\to \infty} \sqrt[k]{|a_k|}>1</math>. So if the ratio test is applicable, then the root test is applicable.

The converse is not true, as the above example shows. We will dispense here with the somewhat theoretical and lengthy proof of the inequality. Advanced students are welcome to try solving the corresponding [[Serlo:_EN:_Exercises:_Subsequences,_Accumulation_points_and_Cauchy_sequences|Exercises]].

== Raabe's criterion ==

In case the ratio test in the above form fails because, for example, <math>\lim_{k\to \infty} \left| \tfrac{a_{k+1}}{a_k} \right| = 1</math>, there is a tightened form where one has to estimate the quotient sequence <math>\left( \left| \tfrac{a_{k+1}}{a_k} \right| \right)_{k \in \N}</math> more precisely. It is called Raabe's criterion and is an [[w:EN:Ratio_test#Extensions_for_L_=_1|extension for ratio 1]] of the ratio test. The name goes back to the Swiss mathematician [[w:en:Joseph Ludwig Raabe|Joseph Ludwig Raabe]]. 

Raabe's criterion is often not as easy to apply as the ratio test and is often not covered in basic lectures. Therefore, we only mention it here and refrain from a derivation. For advanced students who want to derive the criterion, we recommend the corresponding [[Serlo:_EN:_Exercises:_Convergence_criteria_for_series|Exercise]]. Raabe's criterion reads as follows

{{:Math for Non-Geeks: Template:Satz
 |titel=Raabe's criterion
 |satz=
# If <math>\left| \frac{a_{k+1}}{a_k} \right| \le 1-\frac{\beta}{k+1}</math> for almost all <math>k \in \N</math> with a constant <math>\beta >1</math>, then the series <math>\sum_{k=1}^\infty a_k</math> converges absolutely.
# If <math>\frac{a_{k+1}}{a_k} \ge 1-\frac{1}{k+1}</math> for almost all <math>k \in \N</math>, then the series <math>\sum_{k=1}^\infty a_k</math> diverges.
 
 |beispiel=
It is very easy to show the divergence of the harmonic series <math>\sum_{k=1}^\infty \frac 1k</math> using Raabe's criterion. Here, <math>a_k=\frac 1k</math> so for all <math>k\in \N</math>, the following applies

{{Math|<math>\begin{align} 
\frac{a_{k+1}}{a_k} & = \frac{k}{k+1} \\[0.5em] 
& = \frac{k+1-1}{k+1} \\[0.5em] 
& = 1- \frac{1}{k+1}
\end{align}</math>}}

So the series diverges.
}}

{{:Math for Non-Geeks: Template:Beispiel
 |titel=Raabe's criterion for convergence
 |beispiel=
It is somewhat more difficult to prove the convergence of the series <math display="inline">\sum_{k=1}^\infty \frac 1{k^2}</math>. For any <math>n \in \N</math>:

{{Math|<math>\begin{align} 
\frac{a_{k+1}}{a_k} & = \frac{k^2}{(k+1)^2} \\[0.5em] 
& = \left( \frac{k+1-1}{k+1} \right)^2 \\[0.5em] 
& = 1- \frac{2}{k+1} + \frac{1}{(k+1)^2}\\[0.5em] 
& = 1- \frac{2(k+1)-1}{(k+1)^2}\\[0.5em] 
& = 1- \frac{\frac{2(k+1)-1}{k+1}}{k+1}\\[0.5em] 
& = 1- \frac{2-\frac{1}{k+1}}{k+1}\\[0.5em] 
& \left\downarrow\ \begin{array}{rl}
k+1 \ge 2 &\implies \frac{1}{k+1} \le \frac 12 \\
&\implies -\frac{1}{k+1} \ge -\frac 12 \\
&\implies 2-\frac{1}{k+1} \ge 2-\frac 12 = \frac 32 \\
&\implies 1-\frac{2-\frac{1}{k+1}}{k+1} \le 1-\frac{\frac 32}{k+1}
\end{array} \right.\\[0.5em] 
& \le 1-\frac{\frac 32}{k+1}. 
\end{align}</math>}}

With <math>\beta = \frac 32 >1</math> we have that <math>\left| \frac{a_{k+1}}{a_k}\right| \le 1-\frac{\beta}{k+1}</math>. So the series converges absolutely.
}}

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