rewrite $\sum_{n=0}^\infty \frac{(-1)^nx^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$ Rewrite $\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$.
These two series are equal because they can both be derived from $\frac{1}{x^2+2x}$. To get the first one, you split it into $\frac{1}{x}\cdot \frac{1}{x+2}$ and then you use the power series. To get the second series, you complete the square with $\frac{1}{x^2+2x+1-1}$.
The point is those two are obviously equal but I can't figure out how to manipulate them while leaving them in series form to make them both equal each other (without using $\frac{1}{x^2+2x}$ as an in between equals). This was a problem that came up while my teacher was lecturing and he couldn't figure it out.
 A: One tricky part is that you have the singularity at $x=0$ built into the left side, but not the right.
So you'll need to use that $\frac{1}{x}=\frac{-1}{1-(1+x)}=-\sum (1+x)^k$.
These two are equal only on the interval $(-2,0)$ since the left side converges on $(-2,2)\setminus\{0\}$ and the right side converges on $(-2,0)$. 
Writing $$f(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$$
We rewrite it as: $$\begin{align}f(x)
&=\frac{1}{2x}+\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2^{n+1}}\\
&=\frac{1}{2x}+\sum_{m=0}^{\infty}\frac{(-1)^{m+1}x^m}{2^{m+2}}\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^mx^m\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^m\left(1+x-1\right)^m\\
&=\frac1{2x}-\frac14\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^m\sum_{k=0}^{m}\binom{m}{k}(1+x)^k(-1)^{m-k}\\
&=\frac{1}{2x}-\frac14\sum_{k=0}^{\infty}\sum_{m=k}^{\infty}\binom{m}{k}(1+x)^k(-1)^k\left(\frac 12\right)^m\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{k=0}^{\infty}(-1-x)^k\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}\\
\end{align}$$
I'll prove below: $$\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}=2.$$
So you get:
$$f(x)=\frac{1}{2x}-\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k(1+x)^k$$
Also, $$\frac{1}{2x}=\frac{-1}2\frac1{1-(1+x)}=\frac{-1}{2}\sum_{k=0}^{\infty}(1+x)^k$$
Now $$(1+x)^k+(-1)^k(1+x)^k=\begin{cases}2(1+x)^k&k\text{ even}\\0&k\text{odd}\end{cases}$$
So you get: $$f(x)=-\sum_{k=0}^{\infty}(1+x)^{2k}$$

Lemma: For integer $k\geq0$, we have $$\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}=2.$$
Proof: We'll use a probability proof.
Flip a fair coin repeatedly, and stop when you have exactly $k+1$ heads. Let $X$ be the random variable equal to the number of tosses you needed.
Then $P(X=m+1)=\binom{m}{k}\frac{1}{2^{k+1}}$, because you must get exactly $k$ heads in the first $m$ tosses, and then the next toss must be a heads.
But then $\sum_{m=0}^{\infty}P(X=m+1)=1$. Multiplying by $2$ gives us our lemma.
[To deduce that last step, you technically also have to prove that $P(X<\infty)=1$ - that is, that is, there is zero probability that you never get $k+1$ heads.]

If you instead want to prove that:
$$\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^{n+1}}=-x\sum_{n=0}^{\infty}(1+x)^{2n}$$ when both are defined, you can do so fairly easily by noting that:
$$\begin{align}-x\sum_{n=0}^\infty(1+x)^{2n}&=(1-(1+x))\sum_{n=0}^{\infty}(1+x)^{2n}\\
&=\sum_{n=0}^{\infty}(1+x)^{2n}-\sum_{n=0}^{\infty}(1+x)^{2n+1}\\
&=\sum_{m=0}^{\infty}(-1)^m(1+x)^{m}
\end{align}$$
Then a variant of the proof above proof converts $xf(x)$ to this same series:
$$\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^{n+1}}&=
\sum_{n=0}^{\infty}\frac{(-1)^n(1+x-1)^n}{2^{n+1}}\\
&=\sum_{k=0}^{\infty}(1+x)^k(-1)^k\sum_{n=k}^{\infty}\binom{n}{k}\frac{1}{2^{n+1}}\\
&=\sum_{k=0}^{\infty}(-1)^k(1+x)^k
\end{align}$$
A: $\sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{2^{n+1}}
=\sum_{n=0}^\infty (-1)^n(x/2)^{n+1}
$
converges for
$-2 \lt x \lt 2$
and
$-\sum_{n=0}^\infty (x+1)^{2n}
$
converges for
$-1 < x+1 < 1$
or
$-2 < x < 0$.
They are representations of
$\frac{1}{x^2+2x}
$
in different regions.
You can get another representation by
$\frac{1}{x^2+2x}
=\frac1{x^2}\frac{1}{1+2/x}
=\frac1{x^2}\sum_{n=0}^{\infty} (-1)^n (2/x)^n
$
which converges for
$|x| > 2$.
Look up analytic continuation
for fun and profit.
