Proof of an infinite sum using Fourier Series I was revising for my calculus exam and I came across a question that asked to find the Fourier Series of $f(x)=1+x$, on $-1<x<1$, which I did. Which I found to be:
$$f(x) = 1+\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x)$$
I checked this by graphing it and it seemed to hold. However, the latter part of the question asks to evaluate the FS as $x=\frac{1}{2}$, to calculate:
$$\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{2m-1}$$ 
When I set $x=\frac{1}{2}$, I end up with:
$$\frac{3}{2} = 1+\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n\pi} \sin\bigg(\frac{n\pi}{2}\bigg)$$
$$\frac{\pi}{4} = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} \sin\bigg(\frac{n\pi}{2}\bigg)$$
However I cannot transform the sine function into something else, any help?
 A: HINT$1$: the sign at the sum is wrong, because the derivative at $x=0$ must be positive.
HINT$2$: 
$$\sin\frac\pi2n = 
\begin{cases}
0, \text{ if }n=2k\\
(-1)^k, \text{ if }n=2k+1\\
\end{cases}$$
HINT$3$:
$$\frac\pi4 = \arctan 1 = \sum_0^\infty (-1)^k\frac1{2k+1}.$$
A: If you look at the sine graph then you'll see that
$$\begin{eqnarray*}
\sin(0) &=& 0 \\ \\
\sin\left(\frac{\pi}{2}\right) &=& +1  \\ \\
\sin\left(\frac{2\pi}{2}\right) &=& 0 \\ \\
\sin\left(\frac{3\pi}{2}\right) &=& -1 \\ \\
\sin\left(\frac{4\pi}{2}\right) &=& 0 \\ \\
\sin\left(\frac{5\pi}{2}\right) &=& +1
\end{eqnarray*}$$
This pattern continues because $\sin(x+2\pi) \equiv \sin(x)$ for all real $x$. What we see is that for all integers $k \ge 1$, we have
$$\sin\left(\frac{n\pi}{2}\right) = 
\left\{ \begin{array}{ccc}
(-1)^{k+1} & : & n =2k-1 \\
0 & : & \mathrm{otherwise}
\end{array}\right.$$
This means you can ignore all of the terms in your sequence for which $n$ is even, and only consider the terms for which $n$ is odd. This gives:
$$\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\,\sin\left(\frac{n\pi}{2}\right) &=& \sum_{k=1}^{\infty} \frac{(-1)^{(2k-1)+1}}{(2k-1)}\,\sin\left(\frac{(2k-1)\pi}{2}\right) \\ \\
&=& \sum_{k=1}^{\infty} \frac{(-1)^{2k}(-1)^{k+1}}{(2k-1)} \\ \\
&=& \sum_{k=1}^{\infty} \frac{(-1)^{3k+1}}{(2k-1)} \\ \\
&=& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2k-1)} 
\end{eqnarray*}$$
In the last step, I used the fact that, for all $k \in \mathbb N$,
$$(-1)^{3k+1}=(-1)^{2k}\cdot(-1)^{k+1} = \left[(-1)^2\right]^k \cdot (-1)^{k+1} = [+1]^k\cdot (-1)^{k+1} = (-1)^{k+1}$$
