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.

$\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.
