# 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.

• Do you want to show that it converges (in which case a comparison with $\sum_n \frac 1{3^{2n}}$ will do) or compute the value?
– Ant
Apr 5, 2017 at 20:44
• I want to compute the exact value Apr 5, 2017 at 20:46
• I see; I was confused because you mentioned the root test, which can help estabilish if a series converges or not but not its value :)
– Ant
Apr 5, 2017 at 20:48

HINT:

$$\int_0^{1/\sqrt{3}} t^{4n}\,dt=\frac1{\sqrt{3}}\frac{1}{(4n+1)3^{2n}}$$

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?

• play the usual game? Apr 5, 2017 at 20:50
• @Herrpeter yes, differentiate termwise, integrating termwise, re-index etc and then plug in $x=1/3$. Apr 5, 2017 at 20:51
• Lol, the usual game! I love it! Apr 5, 2017 at 20:54
• @Herrpeter also note, the root test is not a good place to go when you are trying to compute a series, it only (sometimes) tells you whether some (absolutely convergent) series converge. It occurred to me this might be the source of the incongruence between your title and body Apr 5, 2017 at 21:09

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.