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.
 A: @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.
A: I know why the Olivier's solution gives us a contradiction. $$\sum_{1}^{\infty}(\frac{1}{i}-\frac{1}{2i+1}-\frac{1}{2i-1})=\sum_{1}^{\infty}\int dx(x^{i-1}-x^{2i}-x^{2i-2}) $$,
but we cannot just exchange the sequence of summation and integral. We have to be care for about it.$$\sum_{1}^{\infty}\int dx(x^{i-1}-x^{2i}-x^{2i-2})=\lim_{n->\infty}\sum_{1}^{n}\int dx(x^{i-1}-x^{2i}-x^{2i-2})$$ .Since finite summation, we can exchange the summation and integral as we wish. 
$$\lim_{n->\infty}\sum_{1}^{n}\int dx(x^{i-1}-x^{2i}-x^{2i-2})=\lim_{n->\infty}\int dx\sum_{1}^{n}(x^{i-1}-x^{2i}-x^{2i-2})=\lim_{n->\infty}\int dx \frac{(1+x)(1-x^n)-x^2(1-x^{2n})-(1-x^{2n})}{1-x^2}$$. For the moment, we cannot just drop the n in the numerator. Use mathematica to calculate the integral, it gives:
$$1+\frac{1}{2n+1}+HarmonicNumber[n]-HarmonicNumber[\frac{1}{2}+n]-log4$$.
Then we can take the limit $n$ goes to $\infty$, we get $1-log 4$.
A: Consider the integral:
$$-\int_{0}^{1}\frac{(1-u)^{2}}{1-u^{2}}du$$
$$=-\int_{0}^{1}\sum_{n=1}^{\infty}\left(u^{2n-2}-2u^{2n-1}+u^{2n}\right)du$$
$$=-\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{n}+\frac{1}{2n+1}\right)$$
Thus, the sum is equal to the integral at the top:
$$=-\int_{0}^{1}\frac{(1-u)^{2}}{1-u^{2}}du = \int_{0}^{1}du-2\int_{0}^{1}\frac{1}{u+1}du=1-2\log(2)$$
