Compute $ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{(4n+1)3^{2n}} $ I think that I should to use the root test, but I still don't know how does it work in this case. I would appreciate any kind of help.
 A: HINT:
$$\int_0^{1/\sqrt{3}} t^{4n}\,dt=\frac1{\sqrt{3}}\frac{1}{(4n+1)3^{2n}}$$
A: Taking the body of your question to be your question, the root test will do nicely
$$
\lim_{n\rightarrow \infty}\frac{1}{(4n+1)^{\frac{1}{n}}3^2}=\frac{1}{9}<1
$$
and the series converges.
To compute it, play the usual 
$$
\sum_{n=1}^{\infty}x^n
$$
games. 
edit: now that I know you are asking for a computation of the series, I will add it for completeness: Note, setting $x=\frac{1}{3}$, your series is
the function 
$$
f(x)=\sum_{n=1}^\infty \frac{x^{2n}}{4n+1}
$$
evaluated at $x=1/3$. Note also,
$$
\frac{x^{4n+1}}{4n+1}=\int x^{4n}dx
$$
which is close to what we want. 
$$
\frac{1}{x}\frac{x^{4n+1}}{4n+1}=\frac{x^{4n}}{4n+1}
$$
And 
$$
f(x^2)=\sum_{n=1}^\infty \frac{x^{4n}}{4n+1}=\frac{1}{x}\int \sum_{n=1}^\infty x^{4n}
$$
can you finish?
A: Positive terms. Criterion of the quotient:
$$ \lim_{n\to \infty}\dfrac{a_{n+1}}{a_{n}}=\lim_{n\to \infty}\dfrac{\frac{1}{(4n+5)3^{2n+2}}}{\frac{1}{(4n+1)3^{2n}}}=\lim_{n\to \infty}\left( \dfrac{4n+1}{4n+5}·\dfrac{1}{3^{2}}\right)=\dfrac{1}{9}<1\ $$
$\Rightarrow$ The serie converges.
