Answer
The interval of convergence is $(-1,1)$ and $f(x)=\frac{2x+1}{1-x^{2}}$.
Work Step by Step
$f(x)=1+2x+x^{2}+2x^{3}+.....$
Consider the power series $g(x)=\Sigma_{k=0}^{\infty}x^{2k}=1+x^{2}+x^{4}+.....$
and $h(x)=\Sigma_{k=0}^{\infty}x^{2k+1}=x+x^{3}+x^{5}+.....$
we observe that $f(x)=g(x)+2h(x)$
$=\frac{2x+1}{1-x^{2}}$
for all the values of $x$ $|x|\lt 1$
Thus, the interval of convergence is $(-1,1)$ and $f(x)=\frac{2x+1}{1-x^{2}}$.
