$\sum_{n=0}^{\infty}(2x^{n}-x^{2n})$ 
Consider $$\sum_{n=0}^{\infty}\left(2x^{n}-x^{2n}\right)$$. Find its set of
  convergence and and function defined as $f(x) =
 \sum_{n=0}^{\infty}(2x^{n}-x^{2n})$ on that set.

As far as I understand, I can consider this series separately, namely
$$\sum_{n=0}^{\infty}(2x^{n}-x^{2n}) = \sum_{n=0}^{\infty}2x^{n} -\sum_{n=0}^{\infty}x^{2n} = 2 \sum_{n=0}^{\infty}x^{n} -\sum_{n=0}^{\infty}x^{2n} = 2 \frac{1}{1-x} - \frac{1}{1-x^2}$$
And the set of convergence to the function above is $|x|<1$.
Am I right?
 A: The general coefficient is 
$$\begin{cases}a_{2n}=1\\a_{2n-1}=2\end{cases}$$
You may use Hadamard's formula:
$$\frac 1r=\limsup_{n\to\infty}\, a_n^{\frac1n}=1.$$
Alternatively, the radius of convergence is the supremum of the $r$ such that $$a_nr^n\xrightarrow[n\to\infty]{}0$$
which is clearly $1$.
A: Yes and no. Your result is correct, but one should exercise a little caution when manipulating infinite series. That is, you must either prove the series is absolutely convergent to rearrange terms as you have, or interpret it formally as a limit of a finite series. The latter is perhaps most instructive:
$$ \begin{split}
\lim_{N\to\infty} \sum_{n=0}^N (2x^n-x^{2n}) &= \lim_{N\to\infty}\left[\frac{2(1-x^{N+1})}{1-x} - \frac{1-x^{2N+2}}{1-x^2}\right] \\
&= \lim_{N\to\infty} \frac{2(1-x^{N+1}+x-x^{N+2})-1+x^{2N+2}}{1-x^2} \\
&= \lim_{N\to\infty} \frac{1+2x - x^{N+1}(2+2x-x^{N+1})}{1-x^2} \\
&= \frac{1+2x}{1-x^2} - \frac{1}{1-x^2}\lim_{N\to\infty} x^N(2+2x-x^N)
\end{split} $$
Now we wave our hands and see if $|x|\geq1$, the limit diverges. If $|x|<1$, then we get the "expected" result. So if we get the same result, why all the extra effort? Infinity can be tough sometimes -- here is a fun link:
https://plus.maths.org/content/when-things-get-weird-infinite-sums
A: \begin{align}
r&=\lim_{n\to \infty}\frac{2x^n-x^{2n}}{2x^{n+1}-x^{2(n+1)}}\\
&=\lim_{n\to \infty}\frac{2-x^{n}}{2x-x^{n+2}}\\
&=\frac{1}{x}\lim_{n\to \infty}\frac{2-x^{n}}{2-x^{n+1}}\\
&=\frac{1}{x}
\end{align}
For convergence we require $|r|>1$ which results in $|x|<1$.
