Mustering $\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2}$ with complex series

Question

I have a series

$$\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2}$$

And I attempted to use the generating function

$$\sum_{k=1}^{\infty} \frac{x^kH_k}{k^2} = \mathrm{Li}_2(1-x)\ln(1-x) + \frac{1}{2}\ln^2(1-x)\ln(x) + \mathrm{Li}_3(x) - \mathrm{Li}_3(1-x) + \zeta(3)$$

In a similar way that joriki did in my other post on another series here: math.stackexchange.com/a/3683523/758843

To start off what I think is right approach, I discerned:

$$\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2} = 4\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{(2k)^2} = 4\sum_{k=1}^{\infty} \frac{(i)^kH_k}{k^2} + 4\sum_{k=1}^{\infty} \frac{(-i)^kH_k}{k^2}$$

I tried using wolfram alpha’s approximation for the sum but it isn’t matching the answer I arrived at.

If I’m wrong, I’d just like to know what I did wrong about breaking the series up into complex series, and not the work for the rest of the problem so that I can do it on my own.

Answer 1

Recall

$$\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}\tag1$$

where if we replace $n$ by $2n$ then multiply both sides by $-\frac{2(-1)^n}{n}$ and sum up over $n$ we have

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^2}=-2\int_0^1 \frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{(-x^2)^n}{n}dx=2\int_0^1\frac{\ln(1-x)\ln(1+x^2)}{x}dx$$

$$\overset{IBP}{=}-2\ln(2)\zeta(2)+4\int_0^1\frac{x\text{Li}_2(x)}{1+x^2}dx$$

$$=-2\ln(2)\zeta(2)+4\int_0^1\frac{x}{1+x^2}\left(-\int_0^1\frac{x\ln(y)}{1-yx}dy\right)dx$$

$$=-2\ln(2)\zeta(2)+4\int_0^1\ln(y)\left(-\int_0^1\frac{x^2}{(1+x^2)(1-yx)}dx\right)dy$$

$$=-2\ln(2)\zeta(2)+4\int_0^1\ln(y)\left(\frac{\pi}{4}\frac{1}{1+y^2}+\frac{\ln(2)}{2}\frac{y}{1+y^2}+\frac{\ln(1-y)}{y}-\frac{y\ln(1-y)}{1+y^2}\right)dy$$

$$=-2\ln(2)\zeta(2)+4\left(-\frac{\pi}{4}G-\frac{\ln(2)}{16}\zeta(2)+\zeta(3)-\int_0^1\frac{y\ln(y)\ln(1-y)}{1+y^2}\right)$$

$$=4\zeta(3)-\frac94\ln(2)\zeta(2)-\pi G-4\int_0^1\frac{y\ln(y)\ln(1-y)}{1+y^2}dy$$

Divide both sides by $4$

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}=\zeta(3)-\frac9{16}\ln(2)\zeta(2)-\frac{\pi}4 G-\int_0^1\frac{y\ln(y)\ln(1-y)}{1+y^2}dy\tag2$$

For the latter integral, differentiate both sides of $(1)$ with respect to $n$ we have

$$\int_0^1 x^{n-1}\ln(x)\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$$

Replace $n$ by $2n$ then multiply both sides by $(-1)^n$ and consider the summation we have

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{2n}+\frac12\ln(2)\zeta(2)=\int_0^1 \frac{\ln(x)\ln(1-x)}{x}\sum_{n=1}^\infty (-x^2)^ndx$$

$$=-\int_0^1\frac{x\ln(x)\ln(1-x)}{1+x^2}dx\tag3$$

Solving $(2)$ and $(3)$ yields

$$\boxed{\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{2n}=\frac{\pi}{4}G+\frac1{16}\ln(2)\zeta(2)-\zeta(3)}$$

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to get your sum, exploit the identity

$$-\ln(1-x)\text{Li}_2(x)=2\sum_{n=1}^\infty\frac{H_n}{n^2}x^n+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}x^n-3\sum_{n=1}^\infty\frac{x^n}{n^3}$$

Replace $x$ by $i$ then consider the real parts of the two sides we have

$$-\Re\{\ln(1-i)\text{Li}_2(i)\}=2\Re\sum_{n=1}^\infty\frac{H_n}{n^2}i^n+\Re\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}i^n-3\Re\sum_{n=1}^\infty\frac{i^n}{n^3}$$

use the fact that $$\Re\sum_{n=1}^\infty i^n f(n)=\sum_{n=1}^\infty (-1)^n f(2n)$$

Thus,

$$-\Re\{\ln(1-i)\text{Li}_2(i)\}=2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{2n}-3\sum_{n=1}^\infty\frac{(-1)^n}{(2n)^3}$$

where $\Re\{\ln(1-i)\text{Li}_2(i)\}=\frac{\pi}{4}G-\frac1{16}\ln(2)\zeta(2)$ and $\sum_{n=1}^\infty\frac{(-1)^n}{n^3}=-\frac34\zeta(3)$

substituting these two results along with the result of $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{2n}$ gives

$$\boxed{\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}=\frac{23}{64}\zeta(3)-\frac{\pi}{4} G}$$

Answer 2

Here we follow the method of Flajolet. Consider the following expansions where n is an integer $$\psi \left( -2z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{2\left( z-n \right)}+{{H}_{2n}}+O\left( z-n \right)n\ge 0$$ $$\psi \left( -2z \right)+\gamma \underset{z\to n-\tfrac{1}{2}}{\mathop{=}}\,\frac{1}{2\left( z-\left( n-\tfrac{1}{2} \right) \right)}+{{H}_{2n-1}}+O\left( z-n+\tfrac{1}{2} \right)n\ge 0$$ $$\psi \left( -2z \right)+\gamma \underset{z\to -n}{\mathop{=}}\,{{H}_{2n-1}}+O\left( z-n \right)n\ge 0$$ $$\frac{\pi }{\sin \left( \pi z \right)}\underset{z\to n}{\mathop{=}}\,{{\left( -1 \right)}^{n}}\left( \frac{1}{\left( z-n \right)}+O\left( z-n \right) \right)$$ Assume $f\left( z \right)$ is an analytic function that goes like $1/{{z}^{2}}$ at infinity and has no other poles in the plane apart from that at $z=0$. From this we have then for $n>0$ $$\frac{\pi \left( \psi \left( -z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right)\underset{z=n}{\mathop{=}}\,{{\left( -1 \right)}^{n}}\left( \frac{f'\left( n \right)}{2\left( z-n \right)}+\frac{{{H}_{2n}}f\left( n \right)}{\left( z-n \right)}+O\left( z-n \right) \right)$$ $$\frac{\pi \left( \psi \left( -z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right)\underset{z=n-\tfrac{1}{2}}{\mathop{=}}\,{{\left( -1 \right)}^{n+1}}\left( \frac{\pi f\left( n-\tfrac{1}{2} \right)}{2\left( z-\left( n-\tfrac{1}{2} \right) \right)}+O\left( 1 \right) \right)$$ $$\frac{\pi \left( \psi \left( -z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right)\underset{z=-n}{\mathop{=}}\,{{\left( -1 \right)}^{n}}\left( \frac{{{H}_{2n-1}}f\left( -n \right)}{\left( z+n \right)}+O\left( z-n \right) \right)$$

The sum over all residues must be zero and so $$\underset{z=n}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right) \right\}=-\underset{z=0}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right) \right\}$$ Using this we find $$\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\left( \frac{1}{2}f'\left( n \right)+{{H}_{2n}}f\left( n \right) \right)}-\sum\limits_{n=1}^{\infty }{\frac{\pi }{2}{{\left( -1 \right)}^{n}}f\left( n-\tfrac{1}{2} \right)}+\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}{{H}_{2n-1}}f\left( -n \right)}=-\underset{z=0}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right) \right\}$$ Recall ${{H}_{2n-1}}={{H}_{2n}}-\frac{1}{2n}$ hence $$\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}{{H}_{2n}}\left( f\left( n \right)+f\left( -n \right) \right)}-\sum\limits_{n=1}^{\infty }{\frac{\pi }{2}{{\left( -1 \right)}^{n}}f\left( n-\tfrac{1}{2} \right)}=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\left( \frac{1}{2n}f\left( -n \right)-\frac{1}{2}f'\left( n \right) \right)}-\underset{z=0}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right) \right\}$$ Assume f is an even function and so obtain

$$\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}{{H}_{2n}}f\left( n \right)}=\frac{1}{4}\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\left( \frac{1}{n}f\left( n \right)-f'\left( n \right)+\pi f\left( n-\tfrac{1}{2} \right) \right)}-\frac{1}{2}\underset{z=0}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right)}f\left( z \right) \right\}$$

For $f\left( n \right)=\frac{1}{{{n}^{2}}}$ therefore $$\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{H}_{2n}}}{{{n}^{2}}}}=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\left( \frac{3}{4{{n}^{3}}}+\frac{\pi }{{{\left( 2n-1 \right)}^{2}}} \right)}-\frac{1}{2}\underset{z=0}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right){{z}^{2}}} \right\}$$ The first series on the right is the zeta function and the second is the definition of Catalan’s constant G. That is $$\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{H}_{2n}}}{{{n}^{2}}}}=-\frac{9}{16}\zeta \left( 3 \right)-\pi G-\frac{1}{2}\underset{z=0}{\mathop{res}}\,\left\{ \frac{\pi \left( \psi \left( -2z \right)+\gamma \right)}{\sin \left( \pi z \right){{z}^{2}}} \right\}$$ Series about $z=0$ yield $$\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{H}_{2n}}}{{{n}^{2}}}}=-\frac{9}{16}\zeta \left( 3 \right)-\pi G-{{\psi }^{\left( 2 \right)}}\left( 1 \right)$$ This has a well-known representation in terms of zeta, i.e. ${{\psi }^{\left( 2 \right)}}\left( 1 \right)=-2\zeta \left( 3 \right)$. The sum is therefore $$\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{H}_{2n}}}{{{n}^{2}}}}=\frac{23}{16}\zeta \left( 3 \right)-\pi G$$

Answer 3

Awesome, so it turns out I just made a few arithmetic errors. The main issue I had was calculating too many complex dilogarithms and logarithms in a page or two in my notebook so I messed up. Also I made an error with the sum.

Let S be the sum I want to calculate.

$$S = 2\bigl(\sum_{k=1}^{\infty} \frac{i^kH_k}{k^2} + \sum_{k=1}^{\infty} \frac{(-i)^kH_k}{k^2}\bigr)$$

The closed form ends up being

$$-G\pi + \frac{\pi^2\ln2}{8} + \frac{29\zeta(3)}{8} - 2\mathrm{Li}_3(1-i) - 2\mathrm{Li}_3(1+i)$$

Where $G$ is Catalan’s constant.