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I got this sum from a site, but I can't remember it.

$$\sum_{j=1}^{\infty}\frac{1}{2j+1}\sum_{i=1}^{\infty}(-1)^i[H_{i,2j}-\zeta(2j)]=\frac{1-\ln(2)}{2}$$

This looking interesting, but is this sum correct?

Let $$A=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(-1)^iH_{i,2j}}{2j+1}$$

$$B=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(-1)^i\zeta(2j)}{2j+1}$$

The sum $A$ and $B$ takes the form $$\sum_{k=0}^{\infty}\frac{(-1)^k F(k)}{2k+1}$$ As for sum $B$ for $F(k)=1$ it resembles $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}$$

1 Answers1

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If $a_n>0$ and $\displaystyle\sum_{n=1}^{\infty}a_n$ converges, then $\displaystyle\sum_{k=1}^{\infty}(-1)^{k-1}\sum_{n=k+1}^{\infty}a_n=\sum_{n=1}^{\infty}a_{2n}$. Thus, $$\sum_{j=1}^{\infty}\frac{1}{2j+1}\sum_{i=1}^{\infty}(-1)^i\big(H_{i,2j}-\zeta(2j)\big)=\sum_{j=1}^{\infty}\frac{1}{2j+1}\sum_{i=1}^{\infty}(-1)^{i-1}\sum_{k=i+1}^{\infty}\frac{1}{k^{2j}}\\=\sum_{j=1}^{\infty}\frac{1}{2j+1}\sum_{k=1}^{\infty}\frac{1}{(2k)^{2j}}\color{blue}{=\sum_{k=1}^{\infty}\left(k\ln\frac{2k+1}{2k-1}-1\right)}$$ which is computed here (your answer is correct).

metamorphy
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