How to find the value $\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}}{2n-1}$ converges to? We have been finding the value of sums related to the Harmonic series by using that
$\sum_{n=1}^{N}\frac{1}{n}= ln(N)+\gamma + \varepsilon_N$.
When I try to calculate  $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n-1}$ using that method there is a point where I don´t know how to continue.
I have read that its value is $\frac{\pi}{4}$ but I don´t understand how is that obtained and I would like to know if it´s possible to solve it using the method I know.
This is what I did:
$S_N=\sum_{n=1}^{N}\frac{(-1)^n}{2n-1}$
$S_{2N}=\sum_{n=1}^{2N}\frac{(-1)^n}{2n-1}=-1+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{4N-3}+\frac{1}{4N-1}=$
$=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4N-1}-2(1+\frac{1}{5}+\frac{1}{9}+...+\frac{1}{4N-3})=\sum_{n=1}^{2N}\frac{1}{2n-1}-2\sum_{n=1}^{N}\frac{1}{4n-3}$
.
$\sum_{n=1}^{2N}\frac{1}{2n-1}=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4N-3}+\frac{1}{4N-1}=$
$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4N-1}+\frac{1}{4N}-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4N})=\sum_{n=1}^{4N}\frac{1}{n}-\frac{1}{2}\sum_{n=1}^{2N}\frac{1}{n}$
.
But I don´t know what to do with the second one. I tried to write it like this:
$\sum_{n=1}^{N}\frac{1}{4n-3}=1+\frac{1}{5}+\frac{1}{9}+...+\frac{1}{4N-3}=$
$=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4N-3}+\frac{1}{4N-1}-(\frac{1}{3}+\frac{1}{7}+...\frac{1}{4N-1})=\sum_{n=1}^{2N}\frac{1}{2n-1}-\sum_{n=1}^{N}\frac{1}{4n-1}$
Wich didn´t help because I find $\sum_{n=1}^{N}\frac{1}{4n-1}$ as difficult to solve as $\sum_{n=1}^{N}\frac{1}{4n-3}$ and I have no idea of what else to do.

Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=\sum_{k=0}^n{\frac{(-1)^n}{2n+1}}=\int_0^1{\sum_{k=0}^n{(-x^2)^n}\,dx},$$ and then prove that $$\left|S_n-\int_0^1{\frac{1}{1+x^2}\,dx}\right| \leq n^{-1}.$$ 
