Math for Non-Geeks/ Alternating series test

{{#invoke:Math for Non-Geeks/Seite|oben}} File:Leibniz-Kriterium - Quatematik.webm The alternating series criterion serves to prove convergence of an alternating series, i.e. a series where the pre-signs alternately change from positive to negative, like $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ or $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k{b}_k$ (with all $b_k$ being positive). Series of this kind can be convergent, but not absolutely convergent. I those cases, criteria for absolute convergence will fail, but the alternating series criterion may be successful.

The alternating series criterion goes back to Gottfried Wilhelm Leibniz and was published in 1682.

Introductory example: Convergence of the alternating harmonic series

Series treated by the alternating series criterion will often converge, but not converge absolutely. Perhaps, the most prominent example for such a converging, but not absolutely convergent series is the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}$. Convergence of it can be shown by making sure that the sequence of partial sums $(S_n{)}_{n\in \mathbb{N}}={\left(\sum _{k=1}^n\frac{(-1{)}^{k+1}}{k}\right)}_{n\in \mathbb{N}}$ converges. For $n=1,2,3,4,5,6,7,8$ , those partial sums are

Those partial sums tend to make jumps of ever smaller getting distance. Those partial sums with odd indices, ($S_{2n-1}$) seem to be monotonically decreasing and those with even indices $S_{2n}$ seem to be increasing. A simple calculation can mathematically verify this assertion: For all $n\in \mathbb{N}$ there is Template:Math i.e. $S_{2n+1}\le S_{2n-1}$ (monotonically decreasing). Analogously, $S_{2n+2}-S_{2n}=\frac{(-1{)}^{2n+3}}{2n+2}+\frac{(-1{)}^{2n+2}}{2n+1}=-\frac{1}{2n+2}+\frac{1}{2n+1}\ge 0$, so $S_{2n+2}\ge S_{2n}$ (monotonically increasing).

If we could now show that $(S_{2n-1})$ is bounded from below and $(S_{2n})$is bounded from above, then both sequences would converge by the monotonicity criterion. Luckily, this is exacly the case: all odd partial sums are bounded from below by any even partial sum and all even partial sums are bounded from above by any odd partial sum: For all $n\in \mathbb{N}$ Template:Math so $S_{2n-1}\ge S_{2n}$ and $S_{2n}\le S_{2n-1}$. We therefore have the bounds $S_{2n-1}\ge S_{2n}\ge S_2=\frac{1}{2}$ and $S_{2n}\le S_{2n-1}\le S_1=1$. Hence, $(S_{2n-1})$ is bounded from below by $\frac{1}{2}$ and $(S_{2n})$ is bounded from above by $1$ .

The monotonicity criterion now implies that both the subsequences of partial sums $(S_{2n-1})$ and $(S_{2n})$ are convergent.

In order to get convergence of the series, we need to show that the sequence of partial sums $(S_n)$ converges. This is for sure the case, if both the odd and the even subsequence $(S_{2n-1})$ and $(S_{2n})$ converge to the same limit.

How can this be shown? First, let us assign a name too the limits: $\underset{n\to \mathrm{\infty }}{\lim }{S}_{2n-1}=S$ and $\underset{n\to \mathrm{\infty }}{\lim }{S}_{2n}=S^{\prime }$. The statement we want to show can then mathematically be expressed as $S=S^{\prime }$. We show this by subtracting both limits from each other, which is equivalent to taking the limit of the sequence difference: Template:Math Above, we showed $S_{2n-1}-S_{2n}=\frac{1}{2n}$ which is a null sequence: Template:Math Hence, $S-S^{\prime }=0$ which means $S=S^{\prime }$.

From this, we can imply that the sequence of partial sums $(S_n)$ converges to $S$ . Mathematically, we need to stay closer than any $ϵ$ to the limit value after surpassing some sequence element number $N$ for the corresponding $ϵ$ Template:Math For a fixed $ϵ$, both the odd and the even partial sum sequences have a suitable $N$, which we name by $(2N_1-1)$ (odd) and $2N_2$(even): Template:Math After reaching the greater of these two numbers $N=\max \{(2N_1-1),2N_2\}$, both sequences stay closer than $ϵ$ to the limit value and Template:Math

Generalizing the proof idea / alternating series test

Now we consider any alternating series. Can we use the same proof as for the alternating harmonic series to show that our general alternating series converges? The answer will depend on the properties of the general alternating series. We used the following properties from the alternating harmonics series:

• The sequence of coefficients $(b_k)=\left(\frac{1}{k}\right)$ without the alternating presign is monotonically decreasing. This gave us monotonicity and boundedness of the two partial sums $(S_{2n-1})$ and $(S_{2n})$ , so we could show that they converge. Without the monotonicity, this may not be the case.
• Further, we used that $(b_k)=\left(\frac{1}{k}\right)$ is a null sequence. This was needed to show that both $(S_{2n-1})$ and $(S_{2n})$ had the same limit, so $(S_n)$ converges to that limit. If $(b_k)=\left(\frac{1}{k}\right)$ converged to a constant $B$, then $(S_n)$ would in the end tend to "jump" up and down by an amount of $B$ and the limits of $(S_{2n-1})$ and $(S_{2n})$ may differ by $B$, so they are not equal.

No further properties of the alternating harmonic series have been used for the proof. So we may use the above proof steps to show convergence of a general alternating series:

Proof alternative

Alternatively, one may use the Cauchy criterion for proving the alternating series criterion.

Notes to the alternating series test

• Of course, we can also change the series presigns from positive $\to$ negative $\to$ positive $\to ...$ to negative $\to$ positive $\to$ negative $\to ...$ and get a valid convergence criterion for series like $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k{b}_k$. The proof is the same, under an interchange of $(S_{2n-1})$ and $(S_{2n})$.
• One can also start from $k=0$, i.e. consider series like $\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{b}_k$ or $\sum _{k=0}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$. Any starting index $k$ is OK. The proof is just the same including an index shift.
• As above, the alternating series test does only lead convergence, but no absolute convergence. For instance, the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}$ converges by the alternating series test. However, it does not converge absolutely.
• The alternating series test can never be used for implying divergence of a series. If a series fails to meet the criteria for the alternating series test, it can still converge. There is an example warning about this below Template:Noprint

• We could also take $(b_k{)}_{k\in \mathbb{N}}$ to be a non-positive and monotonically increasing null sequence. I.e. it approaches $0$ from below. The proof works the same way. Especially, that means $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges, whenever $(b_k{)}_{k\in \mathbb{N}}$ is just any monotone null sequence.

What, if a condition is not fulfilled?

It is important to check that the 2 conditions for the alternating series test are fulfilled! There are alternating series, which do not meet one. The following examples will illustrate alternating series, where $(b_k)$ is either not converging to 0 (our example converges to 1) or not monotone. Both examples fail to be convergent (although they are alternating). The third example is an alternating series, which fails the alternating series test (as it is not monotone), but nevertheless converges. So the alternating series test does not identify all convergent alternating series.

Conclusion: Error bounds for the limit

The alternating series test can show converges, but does not give us the limit. For instance, for the alternating harmonic series , there is $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}=\ln (2)$ . But this limit can not be computed by the alternating series test. However, we can approximate the limit by considering partial sums and the alternating series test will provide us with a neat upper bound for the error of such an approximation.

We have seen above in this article, that the sequence of partial sums with odd index $(S_{2n-1})$ is monotonically decreasing and converges to the limit $\underset{n\to \mathrm{\infty }}{\lim }{S}_{2n-1}=S$ . Further, $S=\inf \{S_{2n-1}:n\in \mathbb{N}\}$, where the infimum of a set is the greatest possible lower bound to its elements. Hence, $S_{2n-1}\ge S$ for all $n\in \mathbb{N}$, so we have upper bounds for the limit getting better and better. Conversely, $(S_{2n})$ is monotonically increasing with $\underset{n\to \mathrm{\infty }}{\lim }{S}_{2n}=S=\sup \{S_{2n}:n\in \mathbb{N}\}$ . So $S_{2n}\le S$ gives a lower bound for all $n\in \mathbb{N}$. That means, we have an estimate $S_{2n}\le S\le S_{2n-1}$ and $S_{2n}\le S\le S_{2n+1}$.

How good is the estimate? We subtract the two inequalities and get Template:Math So, the series elements serve as a precision indicator for the estimate of the limit by partial sums: Template:Math

Generalizing the alternating series test to the Dirichlet test

The Dirichlet test serves for proving convergence of series of the form $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$ . It extends the alternating series test to cases where there is not $a_k=(-1{)}^{k+1}$ . This is particularly useful, if the presign does not change from element to element (like $+,-,+,-,+,-,+,...$) but can have streaks without a change in between (like $+,-,-,+,-,+,+,...$) . The proof is based on Abel's partial summation, which is quite some work to do. We will not state it here. Math for Non-Geeks: Template:Satz The conditions for $(b_k)$ are exactly the same as for the alternating series test. Actually, with $a_k=(-1{)}^{k+1}$, we just get the alternating series test as a special case: Math for Non-Geeks: Template:Aufgabe

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