Yahir Haas

Answered

2022-01-21

How to prove:

$\sum _{k=1}^{\mathrm{\infty}}\frac{k-1}{2k(1+k)(1+2k)}={\mathrm{log}}_{e}8-2$

Answer & Explanation

sjkuzy5

Expert

2022-01-22Added 11 answers

Let me try. Note that the

$RHS=3\mathrm{ln}2+2=2+3\sum _{k=1}^{\mathrm{\infty}}\frac{{(-1)}^{k+1}}{k}=2+3\sum _{k=1}^{\mathrm{\infty}}\frac{1}{2k-1}-\frac{1}{2k}$

We have

$\sum _{k=1}^{\mathrm{\infty}}\frac{k-1}{2k(k+1)(2k+1)}=\sum _{k=1}^{\mathrm{\infty}}\frac{2k-(k+1)}{2k(k+1)(2k+1)}$

$=\sum _{k=1}^{\mathrm{\infty}}(\frac{1}{(k+1)(2k+1)}-\frac{1}{2k(2k+1)})$

$=\sum _{k=1}^{\mathrm{\infty}}(\frac{2}{2k+1}-\frac{1}{k+1}-\frac{1}{2k}+\frac{1}{2k+1})$

$=2+3\sum _{k=1}^{\mathrm{\infty}}\frac{1}{2k-1}-\frac{1}{2k}$

$=RHS$

We have

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