Alternating versions of Convergent Series summing to half

Question

Using $\zeta(2)$ as an example.

$\sum_{n=0}^\infty \frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6} $

When alternating the values gave -$\frac{\zeta(2)}{2}$

$\sum_{n=0}^\infty \frac{(-1)^n}{n^2}=-1+\frac{1}{2^2}-\frac{1}{3^2}+...=-\frac{\pi^2}{12} $

and changing the negative gave $\frac{\zeta(2)}{2}$

$\sum_{n=0}^\infty \frac{(-1)^n}{n^2}=1-\frac{1}{2^2}+\frac{1}{3^2}-...=\frac{\pi^2}{12} $

This seems to only work on convergent series, since something like the Harmonic Series continues to infinity while the Alternating Harmonic Series converges to ln(2)

So why do the alternating versions produce these values so similar to the non alternating version?

Answer 1

Note that we have in general

$$\begin{align} \sum_{n=1}^{2N} (-1)^na_n&=\sum_{n=1}^N a_{2n}-\sum_{n=1}^Na_{2n-1}\\\\ &=\sum_{n=1}^N a_{2n}-\left(\sum_{n=1}^{2N}a_n-\sum_{n=1}^Na_{2n}\right)\\\\ &=2\sum_{n=1}^N a_{2n}-\sum_{n=1}^{2N} a_n\tag1 \end{align}$$

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We can use $(1)$ to evaluate the series of interest. Using $(1)$, for $a_n=\frac1{n^2}$ we have

$$\begin{align}\sum_{n=1}^{2N}\frac{(-1)^n}{n^2}&=2\sum_{n=1}^N \frac1{(2n)^2}-\sum_{n=1}^{2N}\frac1{n^2}\\\\ &=-\frac12\sum_{n=1}^N \frac1{n^2}-\sum_{n=N+1}^{2N}\frac1{n^2}\tag2 \end{align}$$

Since the first term on the right-hand side of $(2)$ converges to $-\frac{\pi^2}{12}$ while the second term converges to $0$, we obtain the expected result

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$$

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Note that the expression in $(1)$ can be useful in evaluating other series.

EXAMPLE $1$:

For example, in THIS ANSWER, I used $(1)$ with $a_n=\frac{\log(n)}{n}$ to show that

$$\sum_{n=1}^\infty \frac{(-1)^n\log(n)}{n}=\gamma\log(2)-\frac12\log^2(2)$$

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EXAMPLE $2$:

As another example of $(1)$, let $a_n=\frac1n$. Then, we see that

$$\begin{align} \sum_{n=1}^{2N}\frac{(-1)^n}{n}&=2\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^{2N}\frac1{n}\\\\ &=-\sum_{n=N+1}^{2N}\frac1n\\\\ &=-\frac1N\sum_{n=1}^N \frac1{1+(n/N)}\tag3 \end{align}$$

The right-hand side of $(3)$ is the Riemann sum for $-\int_0^1\frac1{1+x}\,dx=-\log(2)$. Therefore, we find that

$$\sum_{n=1}^\infty \frac{(-1)^n}{n}=-\log(2)$$

Answer 2

The alternating series you present is $\eta(2)$, the Dirichlet eta function. More specifically, $$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.$$

It turns out that this series, unlike the zeta series, converges for all $s \gt 0$. We can produce a very special relationship to the zeta function $\zeta(s)$ with some algebraic cleverness:

\begin{align} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + \cdots \\ &= \left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + \cdots \right) - \left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + \cdots \right) \\ &= \left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + \cdots \right) - 2 \left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + \cdots \right) \\ &= \left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + \cdots \right) - 2 \left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + \cdots \right) \\ &= \zeta(s) - 2^{1 - s} \left(1 + 2^{-s} + 3^{-s} + 4^{-s} + \cdots \right) \\ &= \zeta(s) - 2^{1 - s} \zeta(s) \\ &= \left( 1 - 2^{1 - s} \right) \zeta(s). \end{align}

So now, by our new formula $\eta(s) = \left( 1 - 2^{1 - s} \right) \zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.

So if $\zeta(2) = \frac{\pi^2}6$, then $\eta(2) = \left( 1 - 2^{-1} \right) \zeta(2) = \frac12 \frac{\pi^2}6 = \frac{\pi^2}{12},$ as desired.

This formula can also be manipulated into the form $\zeta(s) = \frac{\eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 \lt s \lt 1$.

The reason the series for $\eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $\zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $\ln 2$, but it can be done.)