Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$

Question

Is there an elegant proof for this identity for all real $s \neq -1, -1/2$?

$$\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$

Where $H_x$ represents harmonic numbers.

I found it in a very roundabout way for $s=0,1,2,3,\dots$, (see this answer and this answer) and I don't actually have the proof for the other identity I used:

$$\sum_{n=0}^\infty \sum_{k=1}^{n+1}\frac{H_{k-\frac{1}{2}}}{k} x^n= \frac{2}{x(1-x)} \left( \operatorname{arctanh}^2 \sqrt{x}+\log 2 \log (1-x) \right)$$

So whichever one you can prove works fine for me, but I think the first series should be easier to prove.

Using Ali Shather's comment, we can simplify:

$$\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{2H_{2n}-H_n}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$

One of the interesting consequences:

$$\frac{1}{x} \arcsin^2 \sqrt{x}=\sum_{l=0}^\infty \frac{(4 x)^l}{(2l+1) (l+1) {2l \choose l}}$$

Answer 1

Partial solution to the second double sum:

Lets work on the inner sum first \begin{align} S_n=\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}=\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}+\frac{H_{n+\frac12}}{n+1} \end{align}

by substituting $\ H_{k-\frac12}=2H_{2k}-H_k-2\ln2$, to have

\begin{align} \sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}&=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\sum_{k=1}^{n}\frac{H_k}{k}-2\ln2\sum_{k=1}^{n}\frac{1}{k}\\ &=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\left(\frac{H_n^2+H_n^{(2)}}{2}\right)-2\ln2H_n\\ \end{align}

Therefore $$S_n=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\left(\frac{H_n^2+H_n^{(2)}}{2}\right)-2\ln2H_n+\frac{H_{n+\frac12}}{n+1}\tag{1}$$

Which follows that

$$\sum_{n=0}^\infty\left(\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}\right)x^n=2\underbrace{\sum_{n=0}^\infty\sum_{k=1}^{n}\frac{H_{2k}}{k}x^n}_{\Large S_1}-\underbrace{\sum_{n=0}^\infty\left(H_n^2+H_n^{(2)}\right)x^n}_{\Large S_2}-2\ln2\underbrace{\sum_{n=0}^\infty H_nx^n}_{\Large S_3}+\underbrace{\sum_{n=0}^\infty\frac{H_{n+\frac12}}{n+1}x^n}_{\Large S_4}$$.

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Starting with $S_2$ and by using the following identity: ( proved by SuperAbound here)

$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=0}^\infty \left(H_n^2-H_n^{(2)}\right)x^n\tag{2}$$

Add $\ \displaystyle2\sum_{n=0}^\infty H_n^{(2)}x^n=\frac{2\operatorname{Li}_2(x)}{1-x}$ to both sides of $(2)$, we get $$\boxed{S_2=\frac{2\operatorname{Li}_2(x)+\ln^2(1-x)}{1-x}}$$

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$$\boxed{S_3=-\frac{\ln(1-x)}{1-x}}$$

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and the last sum \begin{align} S_4&=\sum_{n=0}^\infty\frac{H_{n+\frac12}}{n+1}x^n\\ &=\sum_{n=1}^\infty\frac{H_{n-\frac12}}{n}x^{n-1}\\ &=\frac1x\sum_{n=1}^\infty \left(\frac{2H_{2n}-H_n-2\ln2}{n}\right)x^n\\ &=\frac2x\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n-\frac1x\sum_{n=1}^\infty \frac{H_{n}}{n}x^n-\frac{2\ln2}x\sum_{n=1}^\infty \frac{x^n}{n}\\ &=\frac2x\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n-\frac1x(\operatorname{Li}_2(x)+\frac12\ln^2(1-x))-\frac{2\ln2}x(-\ln(1-x))\\ \end{align}

The remaining sum can be simplified as follows:

\begin{align} \sum_{n=1}^\infty \frac{H_{2n}}{n}x^n&=2\sum_{n=1}^\infty \frac{H_{2n}}{2n}(\sqrt{x})^{2n}\\ &=\sum_{n=1}^\infty \frac{H_{n}}{n}(\sqrt{x})^n(1+(-1)^n)\\ &=\operatorname{Li}_2(\sqrt{x})+\frac12\ln^2(1-\sqrt{x})+\operatorname{Li}_2(-\sqrt{x})+\frac12\ln^2(1+\sqrt{x})\\ &=\frac12\operatorname{Li}_2(x)+\frac12\ln^2(1-\sqrt{x})+\frac12\ln^2(1+\sqrt{x}) \end{align}

Thus $$\boxed{S_4=\frac1x\left(\ln^2(1-\sqrt{x})+\ln^2(1+\sqrt{x})-\frac12\ln^2(1-x)+2\ln2\ln(1-x)\right)}$$

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I hope someone will take care of $S_1$ and I hope you find my attempt helpful.

Answer 2

Here is a path for getting

$$ \sum_{\substack{0\le n\\1\le k\le(n+1)}} \frac 1k\cdot H_{k-1/2}\; x^n = \frac{2}{x(1-x)} \Big( \operatorname{arctanh}^2 \sqrt{x} + \log 2 \log (1-x) \Big)\ . $$ (We show this holds as an equality of meromorphic functions in the open unit disk in the complex plane.)

Using $$H_{k-1/2}=(2H_{2k}-H_k)-2\log 2\ ,$$the relation to be shown splits in two parts, we consider them separately. The simpler part is the one with coefficients (in each $x^n$) in $2\log 2\cdot\Bbb Q$, we consider it first, it is a good warm-up. The slightly more complicated part with rationaly coefficients (in $x^n$) is similar, we need to write the Cauchy product of two series, one being again $1/(1-x)=(1+x+x^2+x^3+\dots)$, with more care.

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The simpler series, we substitute $N=n+1\ge1$ explicitly, to make it simpler for the bare eye to follow at one glance: $$ \begin{aligned} \sum_{\substack{0\le n\\1\le k\le(n+1)}} \frac 1k\; x^n &= \frac 1x \sum_{\substack{1\le N\\1\le k\le N}} \frac 1k\; x^N \\ &=\frac 1x\sum_{1\le N}\left(\frac 11+\frac 12+\dots+\frac 1N\right)x^N \\ &=\frac 1x (1+x+\dots+x^N+\dots) \left(\frac x1+\frac {x^2}2+\dots+\frac {x^N}N+\dots\right) \\ &=\frac 1x \cdot\frac 1{1-x} \cdot\Big(\ -\ln(1-x)\ \Big)\ . \end{aligned} $$

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The other part is similar. We isolate the same type of a Cauchy product, one of the factors being $1/(1-x)$. After multiplication with $x$: $$ \begin{aligned} x\sum_{\substack{0\le n\\1\le k\le(n+1)}} \frac 1k(2H_{2k}-H_k)\; x^n &= \sum_{\substack{1\le N\\1\le k\le N}} \frac 1k(2H_{2k}-H_k)\; x^N \\ &\qquad\text{ and we keep this in mind, on the other side,} \\ &\qquad\text{ using $y=\sqrt x$:} \\[3mm] x\cdot\frac{2}{x(1-x)} \operatorname{arctanh}^2 \sqrt{x} &= \frac 2{1-y^2}\left(\sum_{j\ge 1\text{ odd}}\frac 1j y^j\right)^2 \\ &= \frac 2{1-y^2}\sum_{2\le M\text{ even}} \sum_{\substack{ j\ge 1\text{ odd}\\ k\ge 1\text{ odd}\\ j+k = M }}\frac 1j\cdot \frac 1k \; y^M \\ &= 2\Big(1+y^2+y^4+y^6+\dots\Big)\sum_{2\le M\text{ even}} \sum_{\substack{ j\ge 1\text{ odd}\\ k\ge 1\text{ odd}\\ j+k = M }}\frac 1j\cdot \frac 1k \; y^M \\ &= 2 \sum_{2\le M\text{ even}} \sum_{\substack{ j\ge 1\text{ odd}\\ k\ge 1\text{ odd}\\ j+k \le M }} \frac 1j\cdot \frac 1k \; y^M \text{ and with $M=2N$, $x=y^2$,} \\ &= 2 \sum_{1\le N\text{ arbitrary}} \sum_{\substack{ j\ge 1\text{ odd}\\ k\ge 1\text{ odd}\\ j+k \le 2N }} \frac 1j\cdot \frac 1k \; x^N \ . \end{aligned} $$ Let us try now to figure out the simplification scheme for the two expressions of the coefficients in $x^N$, $N$ integer $\ge 1$, for a fixed small value of $N$. We take $N=5$. The sum over the $\frac 1j\cdot \frac 1k$ with odd $j,k\ge 1$, having sum $\le 2N$ has the terms explicitly: $$ \text{twice}\qquad \begin{array}{ccccc} \frac 11\cdot \frac 11 & \frac 11\cdot \frac 13 & \frac 11\cdot \frac 15 & \frac 11\cdot \frac 17 & \color{magenta}{\frac 11\cdot \frac 19 } \\ \frac 13\cdot \frac 11 & \frac 13\cdot \frac 13 & \frac 13\cdot \frac 15 & \color{magenta}{\frac 13\cdot \frac 17} & \\ \frac 15\cdot \frac 11 & \frac 15\cdot \frac 13 & \color{magenta}{\frac 15\cdot \frac 15} & & \\ \frac 17\cdot \frac 11 & \color{magenta}{\frac 17\cdot \frac 13} & & & \\ \color{magenta}{\frac 19\cdot \frac 11} & & & & \end{array} $$ The other side is $$ \begin{aligned} \sum_{1\le k\le N} \frac 1k(2H_{2k}-H_k) =& \frac 11\left(\frac 21\color{blue}{+\frac 22-\frac 11}\right) \\ +& \frac 12\left(\frac 21\color{blue}{+\frac 22}+\frac 23\color{blue}{+\frac 24-\frac 11-\frac 12}\right) \\ +& \frac 13\left(\frac 21\color{blue}{+\frac 22}+\frac 23\color{blue}{+\frac 24}+\frac 25\color{blue}{+\frac 26-\frac 11-\frac 12-\frac 13}\right) \\ +& \frac 14\left(\frac 21\color{blue}{+\frac 22}+\frac 23\color{blue}{+\frac 24}+\frac 25\color{blue}{+\frac 26}+\frac 27\color{blue}{+\frac 28-\frac 11-\frac 12-\frac 13-\frac 14}\right) \\ +& \frac 14\left(\frac 21\color{blue}{+\frac 22}+\frac 23\color{blue}{+\frac 24}+\frac 25\color{blue}{+\frac 26}+\frac 27\color{blue}{+\frac 28}+\frac 29\color{blue}{+\frac 2{10}-\frac 11-\frac 12-\frac 13-\frac 14-\frac 15}\right) \\[3mm] =& \frac 11\left(\frac 21\right) \\ +& \frac 12\left(\frac 21+\frac 23\right) \\ +& \frac 13\left(\frac 21+\frac 23+\frac 25\right) \\ +& \frac 14\left(\frac 21+\frac 23+\frac 25+\frac 27\right) \\ +& \color{magenta}{ \frac 15\left(\frac 21+\frac 23+\frac 25+\frac 27+\frac 29\right)} \\[3mm] \ . \end{aligned} $$ So inductively we would have to extrapolate the role of the magenta terms, then show their equality inductively, explicitly: $$ \color{magenta}{ \sum_{\substack{1\le j,k\text{ odd}\le 2N\\j+k=2N}} \frac 1j\cdot\frac 1k = \frac 1N\sum_{1\le j\text{ odd}\le 2N}\frac 1j \ .} $$ Showing this inductively would break the harmonic beauty, so let us do the job thematically. And this again in the special case $N=5$, since the idea can be easily extrapolated. (It is a typical MZV (multiple zeta values) idea.) The integral form for the equality to be shown is (with $a,b$ playing the roles of $j-1$, $k-1$): $$ \sum_{\substack{0\le a,b\text{ even}\le 2N-2\\a+b=2N-2}} \iint_{[0,1]^2}x^a\; y^b\; dx\; dy = \frac 2{2N} \sum_{0\le a\text{ even}\le 2N-2}t^a\; dt \ . $$ The idea is now to write the double integral on the square as twice the double integral on the simplex $0\le x\le y\le 1$, then change variables, $y=Y$, $x=tY$ with $t,Y\in[0,1]$. Then (formally) $dx\wedge dy = d(tY)\wedge dY=Y\; dt\wedge dY$, and we get a uniform factor $Y^{a+b+1}=Y^{2N-1}$, and by integration the factor $\frac1{2N}$.

We have now completed also the second part.

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This became a long answer, but i hope the combinatorial idea behind the steps became in this way easy and transparent.