Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$

Question

How to prove that

$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})} {512}$$

This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.

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Any way, I tried hard on this one but no success yet. here is what I did:

Using the identity

$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$

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

We can write

$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$

where

\begin{align} \sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\ &=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\ &=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\ &=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right) \end{align}

Therefore

$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$

The sum can be done using the following identity

$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$

Differentiate both sides with respect to $a$ then set $a=1/2$ we get

$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$

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and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks

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Answer 1

Cornel's way to make it easy. Replace the harmonic number in the numerator by Digamma function, using that $H_{n/2}= \psi(n/2+1)+\gamma$, and then splitting the series using the parity, we have

$$ S=\sum_{n=1}^{\infty} \frac{ \psi(n/2+1)+\gamma}{(2n+1)^3}=\sum_{n=1}^{\infty} \frac{ \psi(n+1)+\gamma}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ \psi(n+1/2)+\gamma}{(4n-1)^3}$$ $$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ 2H_{2n}-H_n-2\log(2)}{(4n-1)^3}$$ $$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}-\sum_{n=1}^{\infty} \frac{H_n}{(4n-1)^3}-2\log(2)\sum_{n=1}^{\infty} \frac{1}{(4n-1)^3}+2\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3},$$ and since the first two series are straightforward using Cornel's Master Theorem of Series from A master theorem of series and an evaluation of a cubic harmonic series, which is also given in the book, (Almost) Impossible Integrals, Sums, and Series, and then noting that

$$\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3}=\frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{H_{n}}{(2n-1)^3}-\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n-1)^3}\right),$$

where for the first series we can use the same mentioned master theorem, and then the second one is already known in the form $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n+1)^3}$ (it's easy to rearrange the series according to our needs), and you may find its value here together with a solution in comments, we're done.

End of story.

Answer 2

Using W.A.as well as my previous calculations on harmonics sums. I find $$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad -\frac{\pi^2G}{4}-2G+\frac{7}{4}\zeta(3)-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi}{2}-\frac{\pi^2}{4}+\frac{\pi^3}{16}+\frac{\pi^4}{64}+\ln2-2-3\beta(4)+\frac{\Psi^{(3)}(\frac{1}{4})}{256}- \frac{\Psi^{(3)}(\frac{3}{4})} {256}$$ $$ S=0,047743102114778065267...$$ But I'm not sure of the result.

Answer 3

Using the identity ( I can provide the proof if wanted but it's a nice problem to try)

$$\sum_{n=0}^\infty (-1)^n\left(H_{n/2}-H_n+\ln2\right)\cos(x(2n+1))=\frac{x}{2\sin x}$$

Multiply both sides by $x^2$ then integrate from $x=0$ to $\pi/2$ to get

\begin{align} I&=\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx=\sum_{n=0}^\infty (-1)^n\left(H_{n/2}-H_n+\ln2\right)\int_0^{\pi/2}x^2\cos(x(2n+1))\\ &=\sum_{n=0}^\infty (-1)^n\left(H_{n/2}-H_n+\ln2\right)\left(\frac{\pi^2}{4}\frac{\cos(nx)}{2n+1}-\frac{2\cos(nx)}{(2n+1)^3}-\frac{\pi\sin(nx)}{(2n+1)^2}\right) \end{align}

Note that inside the sum, $\cos(nx)$ and $\sin(nx)$ behave as $(-1)^{n}$ and $0$ respectively, then

$$I=\frac{\pi^2}{4}\sum_{n=0}^\infty\frac{H_{n/2}-H_n+\ln2}{2n+1}-2\sum_{n=0}^\infty\frac{H_{n/2}}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{H_{n}}{(2n+1)^3}-2\sum_{n=0}^\infty\frac{\ln2}{(2n+1)^3}$$

rearrange

$$\sum_{n=0}^\infty\frac{H_{n/2}}{(2n+1)^3}=\sum_{n=0}^\infty\frac{H_{n}}{(2n+1)^3}-\underbrace{\sum_{n=0}^\infty\frac{\ln2}{(2n+1)^3}}_{\frac78\ln2\zeta(3)}+\frac{\pi^2}{8}\underbrace{\sum_{n=0}^\infty\frac{H_{n/2}-H_n+\ln2}{2n+1}}_{M}-\frac12I$$

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From the question body, we have

$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$

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To calculate $M$, we use the identity

$$\int_0^1\frac{x^n}{1+x}\ dx=H_{n/2}-H_n+\ln2$$

\begin{align} M&=\int_0^1\frac{1}{1+x}\sum_{n=0}^\infty\frac{x^n}{2n+1}\ dx=\int_0^1\frac{\tanh^{-1}\sqrt{x}}{\sqrt{x}(1+x)}\ dx\\ &=2\int_0^1\frac{\tanh^{-1}x}{1+x^2}\ dx=-\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}\ dx=-\int_0^1\frac{\ln x}{1+x^2}\ dx=G \end{align}

.

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\begin{align} I&=\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx\overset{IBP}{=}-\frac32\int_0^{\pi/2} x^2 \ln(\tan(x/2))\ dx\\ &=3\sum_{n=0}^\infty \frac{1}{2n+1}\int_0^{\pi/2} x^2 \cos(x(2n+1))\ dx\\ &=3\sum_{n=0}^\infty \frac{1}{2n+1}\left(\frac{\pi^2}{4}\frac{(-1)^n}{2n+1}-\frac{2(-1)^n}{(2n+1)^3}\right)\\ &=\frac{3\pi^2}{4}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}-6\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}\\ &=\frac{3\pi^2}{4}G-6\beta(4) \end{align}

where $\beta(4)$ $=\frac1{768}\psi_3(1/4)-\frac{\pi^4}{96}$

By combining these results we get the desired closed form.

Answer 4

I asked Cornel for a solution to the nice key result from Ahmad Bow's solution. Here is a solution in large steps. We need two known results, that is $\displaystyle \int_0^1\frac{x^n}{1+x}\textrm{d}x=H_{n/2}-H_n+\log(2)$ and $\displaystyle \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$. If replacing $p$ by $i \sqrt{p}$ in the last series, make rearrangements and reindexing the series, we obtain$\displaystyle \sum _{n=0}^{\infty } (-1)^n y^n \cos ((2 n+1) x)=\frac{(1+y) \cos (x)}{1+2y \cos (2 x)+y^2}$.

Then, we have \begin{equation*} \sum_{n=0}^\infty (-1)^n\left(H_{n/2}-H_n+\ln2\right)\cos((2n+1)x)=\int_0^1\frac{1}{1+y}\sum_{n=0}^\infty (-1)^n y^n\cos((2n+1)x)\textrm{d}y \end{equation*} \begin{equation*} =\cos (x)\int_0^1 \frac{1}{1+2y \cos (2 x)+y^2}\textrm{d}y=\cos (x)\int_{\cos(2x)}^{1+\cos(2x)} \frac{1}{t^2+\sin^2(2x)}\textrm{d}t=\frac{x}{2\sin(x)}. \end{equation*}

End of story.

Answer 5

Another proof besides Cornel's one:

Let

\begin{align} S&=\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)e^{ix(2n+1)}\\ &=\sum_{n=0}^\infty(-1)^n\int_0^1\frac{y^n \ dy}{1+y}e^{ix(2n+1)}\\ &=\int_0^1\frac{e^{ix}\ dy}{1+y}\sum_{n=0}^\infty\left(-ye^{2ix}\right)^n\\ &=\int_0^1\frac{e^{ix}\ dy}{(1+y)(1+e^{2ix}y)}\\ &=\frac{e^{ix}}{1-e^{2ix}}\left(\ln2-\ln(1+e^{2ix})\right)\\ &=\frac{i}{2\sin x}\left(-\ln(\cos x)-ix\right)\\ &=\frac{x}{2\sin x}-i\frac{\ln(\cos x)}{2\sin x} \end{align}

substituting $e^{ix(2n+1)}=\cos(x(2n+1))+i\sin(x(2n+1))$ and comparing the real and imaginary parts we get:

$$\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)\cos(x(2n+1))=\frac{x}{2\sin x}\tag1$$

$$\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=-\frac{\ln(\cos x)}{2\sin x}\tag2$$

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If we follow the same approach, we get

$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\cos(x(2n+1))=-\frac{\ln(\sin x)}{2\cos x}\tag3$$

$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=\frac{\pi/2-x}{2\cos x}\tag4$$

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The identity in (1) is discovered by @Ahmad Bow.