calculate the sum of an infinite series Here is the series:
$$
\sum_{k = 0}^{\infty} \frac{(-1)^k}{2k + 1}.
$$
I don't know how to start at all. Thank you for your help. 
 A: Let $S=\sum_{k=0}^\infty\dfrac{x^{2k+1}}{2k+1}$
$$\dfrac{dS}{dx}=\sum_{k=0}^\infty(x^2)^k=\dfrac1{1-x^2}$$  for $|x^2|<1$
Integrate both sides to get $$S=\ln\dfrac{1+x}{1-x}+K$$
$x=0\implies0=\ln1+K\iff K=0$
A: $$\dfrac{(-1)^k}{2k+1}=-i\cdot\dfrac{i^{2k+1}}{2k+1}$$
Now for $-1\le x<1,$  $$\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\cdots$$
$$\ln(1+x)-\ln(1-x)=?$$
A: I'm a student, so I may not have the mathematical rigour necessary, but I'll try to help.
It turns out that the Taylor Series for $\arctan(x)$ is this:
$$\arctan(x) = \sum_{k=0}^\infty (-1)^k\frac{(x^{2k+1})}{2k+1} = \frac{x}{1}-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$$
So if x=1, you get:
$$\arctan(1) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$$
Which is exactly your series. Since $\arctan(1)=\frac{\pi}{4}$, your series converges to $\frac{\pi}{4}$. So
$$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = \frac{\pi}{4}$$
Some other things:
How do I know your series converges? Well, because of Leibniz's rule for alternating series, which you'll find here.
How do I know that's the Taylor Series for $\arctan(x)$? If you take the derivative of $\arctan(x)$, which is $\frac{1}{1+x^2}$, and now you take its Taylor series  and integrate it, you get the Taylor Series for $\arctan(x)$. Here's a Quora Post that explains how to obtain it:
(The links are spaced out because Quora won't let me otherwise)
