Is my proof that $\frac{\pi}{4}=\sum\limits_{n\geq 0}(-1)^n \frac{1}{2n+1}$ correct? Respected All
I was trying to prove that 
$$\sum_{n\geq 0}(-1)^{n} \frac{1}{2n+1}=\frac{\pi}{4}$$
What I tried to show like this.
We know 
$$\frac{1}{1+x^2}=(1+x^2)^{-1}=\sum_{n\geq 0}(-1)^nx^{2n}, |x^2|<1$$
Integrating term by term with in interval of convergence we get 
$$\int_0^{x_0}\frac{dx}{1+x^2}=\sum_{n\geq 0}(-1)^n\frac{x_0^{2n+1}}{2n+1}$$
Taking $x_0=1$ we have 
$$[\arctan(x)]_0^1=\sum_{n\geq 0}(-1)^n \frac{1}{2n+1}$$
i.e. $$\frac{\pi}{4}=\sum_{n\geq 0}(-1)^n \frac{1}{2n+1}$$
Am I right?
 A: You are already 90% of the way there; using term-by-term integration or any of the other methods described at Taylor series of $\arctan$, you can derive
$$\arctan x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1},\qquad |x|<1,$$
but you cannot claim directly that this is true for $x=1$ because this is out of the range of the equation. (You can't do term-by-term integration on the interval $[0,1]$ because the $\sum\int=\int\sum$ theorem is only valid when convergence is uniform, and a priori you only know that the power series is uniformly convergent on $[0,x]$ for any $|x|<1$.)
But with a little extra work, we can push this equality out to the boundary of the unit circle, using Abel's theorem. This says that if $\sum_{n=0}^\infty a_n$ is a convergent series then $$\lim_{x\to1^-}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_n.$$
And we know that $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$ is convergent, because it is an alternating series. Thus $$\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}=\lim_{x\to1^-}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}=\lim_{x\to1^-}\arctan x$$ (because the power series is valid for $|x|<1$), and this is equal to $\arctan 1=\frac{\pi}4$ because arctan is continuous in a neighborhood of $1$ (indeed, in a neighborhood of $\Bbb R$).
By the way, just yesterday I submitted a formal proof of this exact theorem in Metamath, so I got to see all the little complications that are glossed over in this exposition.
A: The right hand side
$$S=-i\sum_{n=0}^\infty\dfrac{i^{2n+1}}{2n+1}$$ 
$$2iS=\ln\dfrac{1+i}{1-i}=\ln\dfrac{(1+i)^2}2=\ln(i)$$
using $\ln(1+x)$ expansion
As $i=e^{i\pi/2},$  the principal value of $\ln(i)=\dfrac{i\pi}2$
