$\sum_{n=0}^{\infty}(2x^{n}-x^{2n})$ 
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?

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 $1$.
