Compute : $\sum_{m≥1}\frac{(-1)^{m}}{(m+1)(2m+1)^{2}}$

Question

How can Compute in closed form this double summation :

$\displaystyle\sum_{m≥1}\frac{(-1)^{m}}{(m+1)(2m+1)^{2}}$

I need to evaluate this sum using digamma function

Actually I don't have any ideas to approach it

Answer 1

Actually I don't have any ideas to approach it

Hint. By a partial fraction decomposition, one gets $$ \frac{(-1)^{m}}{(m+1)(2m+1)^{2}}=\frac{(-1)^{m}}{m+1}-\frac{2(-1)^{m}}{2m+1}+\frac{2(-1)^{m}}{(2m+1)^{2}} $$ leading to (use Abel's lemma for example) $$ \sum_{m=1}^\infty\frac{(-1)^{m}}{(m+1)(2m+1)^{2}}=\ln 2-1-\frac{\pi}2+2G $$ where $G$ is Catalan's constant.

Answer 2

As a natural extension of this problem we assume $|z|\le 1$ and calculate the sum

$$s(z) = \sum_{m=1}^\infty \frac{z^m}{(m+1)(2m+1)^2}\tag{1}$$

Oliver Oloa has started with partial fraction decomposition.

Here we adopt a different standard approach which represents the denominator by integrals allowing us to perform the sum easily at the (permissible) expense of having to solve integrals.

We set

$$\frac{1}{(2m+1)^2} = \int_0^1 \left (\frac{1}{x} \int_0^x y^{2m} \,dy \right) \,dx\tag{2}$$

Now interchanging summation and integration we can perform the $m$-sum

$$\sum_{m=1}^\infty \frac{1}{m+1} z^m y^{2m} = \frac{-y^2 z-\log \left(1-y^2 z\right)}{y^2 z}$$

Now the we have to integrate this expression according to

The $y$-integral is

$$i_y(x) = \int_0^x \frac{-y^2 z-\log \left(1-y^2 z\right)}{y^2 z} \, dy = \frac{\log \left(1-x^2 z\right)}{x z}+\frac{2 \tanh ^{-1}\left(x \sqrt{z}\right)}{\sqrt{z}}-x$$

and the subsequent $x$-integral becomes

$$s(z) = \int_0^1 \frac{i_y(x)}{x} \,dx \\ = \frac{2 \left(\text{Li}_2\left(\sqrt{z}\right)-\frac{\text{Li}_2(z)}{4}\right)}{\sqrt{z}}+\frac{-\log (1-z)-2 \sqrt{z} \tanh ^{-1}\left(\sqrt{z}\right)}{z}-1\tag{3}$$

Here's the graph of $s(z)$

Now the harvest

For $z \to -1$ we recover the result given by Olivier Oloa

$$s(-1) = 2 C-\frac{\pi }{2}-1+\log (2) \simeq -0.045718$$

The non-alternating sum is obtained letting $z \to +1$ which gives

$$s(+1) = \frac{1}{4} \left(\pi ^2-4 (1+\log (4))\right) \simeq 0.0811067$$

We might also consider the sum at $\pm \frac{1}{2}$ leading to

$$s(+\frac{1}{2}) = 2 \sqrt{2} \text{Li}_2\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi ^2}{12 \sqrt{2}}-1+\frac{\log ^2(2)}{2 \sqrt{2}}+\log (4)-2 \sqrt{2} \tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right) \simeq 0.0319662$$

$$s(-\frac{1}{2}) = \frac{i \text{Li}_2\left(-\frac{1}{2}\right)}{\sqrt{2}}-2 i \sqrt{2} \text{Li}_2\left(\frac{i}{\sqrt{2}}\right)-1+\log \left(\frac{9}{4}\right)-2 \sqrt{2} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \simeq -0.0249609$$

Here

$$\text{Li}_{n}(z) = \sum _{k=1}^{\infty } \frac{z^k}{k^n}$$

is the polylog function.