<s>[INST] How to find the value of $\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$ Is there any way to solve this summation?
$$\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$$
The value is 1 - log4... but I'm not able to proove it.
 [/INST] @Olivier Oloa, obviously, there is something wrong in your solution.
For i>0, $\frac{1}{i}-\frac{1}{2i+1}-\frac{1}{2i-1}=\frac{1}{2i}-\frac{1}{2i+1}+\frac{1}{2i}-\frac{1}{2i-1}<\frac{1}{2i}-\frac{1}{2i+1}$, but using your method, $\sum_{i=1}^{\infty}\frac{1}{2i}-\frac{1}{2i+1}=\sum\int dx{x^{2i-1}-x^{2i}}=\int dx\frac{x}{1-x^2}-\frac{x^2}{1-x^2}=\int dx \frac{x}{1+x}=1-log2 $. But I cannot see where lies the wrong. And mathematica tells the answer is 1-log4.
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