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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.

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    $\begingroup$ 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? $\endgroup$
    – Ant
    Apr 5, 2017 at 20:44
  • $\begingroup$ I want to compute the exact value $\endgroup$
    – Herrpeter
    Apr 5, 2017 at 20:46
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    $\begingroup$ 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 :) $\endgroup$
    – Ant
    Apr 5, 2017 at 20:48

3 Answers 3

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HINT:

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

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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?

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  • $\begingroup$ play the usual game? $\endgroup$
    – Herrpeter
    Apr 5, 2017 at 20:50
  • $\begingroup$ @Herrpeter yes, differentiate termwise, integrating termwise, re-index etc and then plug in $x=1/3$. $\endgroup$ Apr 5, 2017 at 20:51
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    $\begingroup$ Lol, the usual game! I love it! $\endgroup$ Apr 5, 2017 at 20:54
  • $\begingroup$ @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 $\endgroup$ Apr 5, 2017 at 21:09
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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.

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