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In this link gave an answer for a question which invoved the sum of generalized harmonic numbers with power $2n+1$.

And I thought about this sum: $$S=\sum_{n=1}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+1)^2}$$

I tried to search in our website but still can't find some similiar questions. The "generalized" version can be found in Ali Shadhar's book.

$$S_1=\sum_{n=1}^\infty (-1)^{n}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=-\beta{(4)}+\frac{35\pi}{64}\zeta{(3)}-\frac{\pi^2}{48}G$$

Can someone give me a hint for this sum? Every help is welcomed, thank you very much.

Ali Shadhar
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OnTheWay
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1 Answers1

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In the book you mentioned in page 68, we have

$$\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}x^{n}=\operatorname{Li}_4(x)-2\operatorname{Li}_4(1-x)+2\ln(1-x)\operatorname{Li}_3(1-x)+\frac12\operatorname{Li}_2^2(x)$$ $$-\ln^2(1-x)\operatorname{Li}_2(1-x)-\frac13\ln(x)\ln^3(1-x)+2\zeta(4). $$

Using $$\sum_{n=0}^\infty (-1)^{n}a_{2n+1}=\mathfrak{J}\sum_{n=1}^\infty i^n a_n$$

we have

$$\sum_{n=0}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+1)^2}=\mathfrak{J}\sum_{n=1}^\infty i^{n}\frac{H_n^{2}}{n^2}$$

$$=2\zeta(4)+\mathfrak{J}\{\operatorname{Li}_4(i)-2\operatorname{Li}_4(1-i)+2\ln(1-i)\operatorname{Li}_3(1-i)+\frac12\operatorname{Li}_2^2(i)$$ $$-\ln^2(1-i)\operatorname{Li}_2(1-i)-\frac13\ln(i)\ln^3(1-i)\}$$

Ali Shadhar
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