Math for Non-Geeks/ Root test

{{#invoke:Math for Non-Geeks/Seite|oben}}

Let us turn our attention to the root test, which is a powerful tool for proving the convergence or divergence of a given series. It is based off the direct comparison test, in fact we will compare the series with the geometric series $\sum _{k=1}^{\mathrm{\infty }}{q}^k$ with $0\le q<1$.

Derivation

Recap

We have already learned how to use the direct comparison test with a majorant. In summary, a series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ with $a_k\ge 0$ is convergent, if there exists a convergent series $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ with $a_k\le b_k$.

Furthermore we know that every geometric series $\sum _{k=1}^{\mathrm{\infty }}{q}^k$ with $q\in [0,1)$ is convergent.

First derivation

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ be a series. Also let $a_k\ge 0$ for all $k\in \mathbb{N}$, because this is a constraint we need for the direct comparison test. For our majorant we need a $q\in [0,1)$ with $a_k\le q^k$. Then we have

The series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is convergent by our direct comparison test. The inequality $a_k\le q^k$ can be transformed:

Thus if there exists a $q$ with $0\le q<1$, so that $\sqrt[k]{a_k}\le q$, then $a_k\le q^k$ and the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is convergent.

Bringing limit superior into play

For the convergence behaviour we can ignore the value of finitely many summands. Thus $\sqrt[k]{a_k}\le q$ must not be satisfied for all $k\in \mathbb{N}$, but for all $k\in \mathbb{N}$, with finitely many exceptions. So the inequality $\sqrt[k]{a_k}\le q$ must be satisfied for almost all $k\in \mathbb{N}$.

We can restate the requirement, that there must exist a $q\in [0,1)$ with $\sqrt[k]{a_k}\le q$ for almost all $k\in \mathbb{N}$, using the limit superior:

Or in other terms:

If $\sqrt[k]{a_k}\le q$ for almost all $k$, the succession ${\left(\sqrt[k]{a_k}\right)}_{k\in \mathbb{N}}$ is bounded above and must have an accumulation point less than or equal to $q$. This accumulation point is equal to $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{a_k}$ and $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{a_k}\le q$.

Conversely, let $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{a_k}\le q$ for a $q\in [0,1)$. Then for all all $ϵ>0$ the inequality $\sqrt[k]{a_k}\le q+ϵ$ is satisfied for almost all $k\in \mathbb{N}$. Because $q<1$ there exists $ϵ>0$, that is small enough so that $q+ϵ<1$. Set $\tilde{q}=q+ϵ$. We have $\tilde{q}<1$ and the inequality $\sqrt[k]{a_k}\le \tilde{q}$ holds for almost all $k$.

Summary: Instead of $\sqrt[k]{a_k}\le q$ for almost all $k\in \mathbb{N}$ it suffices to show $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{a_k}<1$, to prove convergence the series.

What about absolute convergence?

What happens if not all $a_k\ge 0$? Then we cannot use the above argumentation, because for even $k$ and $a_k<0$ the values $\sqrt[k]{a_k}$ is not defined. But we still have a valid argumentation if we talk about $\sum _{k=1}^{\mathrm{\infty }}|a_k|$, to show the absolute convergence of the series (from which normal convergence follows immediately). Thus for series with $a_k\ge 0$ for all $k$ we have $|a_k|=a_k$. In this case nothing changes because for $a_k$ we can also use $|a_k|$. The conclusion is:

Root test for divergence

We have derived the root test to show the convergence of a series. Can we also derive a criterion for divergence? Let us assume that $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{|a_k|}>1$. Then for infinitely many $k\in \mathbb{N}$ we have the inequality $\sqrt[k]{|a_k|}\ge 1$. For these $k$ we have $|a_k|\ge 1^k=1$, thus ${(|a_k|)}_{k\in \mathbb{N}}$ cannot be a null sequence. But then ${\left(a_k\right)}_{k\in \mathbb{N}}$ is also no null sequence. From the term test it follows immediately that $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is divergent. We can generalize this case if instead of $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{|a_k|}>1$ we require the inequality $\sqrt[k]{|a_k|}\ge 1$ for almost all $k\in \mathbb{N}$.

Limitations of the root test

In case $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{|a_k|}=1$ we cannot say whether we have convergence or divergence. In fact there exist both convergent and divergent series that satisfy this equation. First consider the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$, which is divergent. We have

But also the convergent series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ satisfies this equation:

These examples show that we cannot conclude convergence or divergence from $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{|a_k|}=1$. In this case we have to employ another convergence criterion!

How to apply the root test

To apply the root test on a series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ we can proceed as follows: We compute $\sqrt[k]{|a_k|}$ and find the limit (if the limit exists) or the limit superior.

1. If $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{|a_k|}<1$, then the series converges absolutely.
2. If $\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[k]{|a_k|}>1$, then the series diverges.
3. If $\sqrt[k]{|a_k|}\ge 1$ for infinitely many $k$, then the series diverges.
4. Else, if none of the above is true, we cannot make a statement about convergence behaviour using the root test.

Exercises

Exercise 2

{{#invoke:Math for Non-Geeks/Seite|unten}}