Does $\sum\limits_{n\geq 1}\frac{1}{(3n-1)(3n+1)}$ converge or diverge? 
How would you prove convergence/divergence of the following series?
$$\sum\limits_{n\geq 1}\dfrac{1}{(3n-1)(3n+1)}$$
  I'm interested in more ways of proving convergence/divergence for this series.

My thoughts:
$$\sum\limits_{n\geq 1}\dfrac{1}{(3n-1)(3n+1)}$$
I'm going to use Integral Test
let $f(n)=\dfrac{1}{(3n-1)(3n+1)}$ 
we note that $f$ is a continuous, positive and decreasing function on the interval $[1,+\infty ($ then 
$$\sum_{n=1}^\infty f(n) \text{ Convergent } \iff \int_1^\infty f(x)\,dx \text{ Convergent }$$
so let's try to compute 
$$\int_1^\infty f(n)\,dn=\int_1^\infty \dfrac{1}{(3n-1)(3n+1)}\,dn  $$
Substitution for $n=\dfrac{1}{3}u$ and $dn=\dfrac{1}{3}du$ we got :
\begin{align}
\int_1^\infty f(n)\,dn &=\int_1^\infty \dfrac{1}{(3n-1)(3n+1)}\,dn  \\
&=\int_1^\infty \dfrac{1}{(3(\dfrac{1}{3}u)-1)(3(\dfrac{1}{3}u)+1)}\,\dfrac{1}{3}du  \\
&=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u-1)(u+1)}\,du  \\
&=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u^2-1)}\,du  \\
\end{align}
note that $\left(arctanh(u)\right)'=\left(\tanh^{-1}(h)\right)'=\dfrac{1}{1-u^{2}}$ so
\begin{align}
\int_1^\infty f(n)\,dn &=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u^2-1)}\,du  \\
&=-\dfrac{1}{3}\int_1^\infty \dfrac{1}{(1-u^2)}\,du  \\
&=-\dfrac{1}{3} \left(\tanh^{-1}(u)\right) \biggl|_{1}^{+\infty} \\
&=-\dfrac{1}{3} \left(\tanh^{-1}(3n)\right) \biggl|_{1}^{+\infty} \\
\end{align}
so can't go ahead beause i don't know the limit of $tanh^{-1}$ i can use woflrame to do that but suppose im in the contest what gonna do 
so my questions:


*

*Is my proof correct

*since the calculation is not easy in this case I'm interested in more ways of proving convergence/divergence for this series.

 A: There is an easier way to check convergence:
$$0\leq \sum_{n\geq 1}\frac{1}{(3n-1)(3n+1)}\leq \sum_{n\geq 1}\frac{1}{8n^2}=\frac{\pi^2}{48}.$$
We may compute the value of the series by considering that:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{(3n-1)(3n+1)}&=&\frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}\left(x^{3n-2}-x^{3n}\right)\,dx\\&=&\frac{1}{2}\int_{0}^{1}\frac{1+x}{1+x+x^2}\,dx\\&=&\color{red}{\frac{1}{2}\left(1-\frac{\pi}{3\sqrt{3}}\right)}.\end{eqnarray*}$$
A: Checking Convergence
Applying a Telescoping Series, we get
$$
\begin{align}
\sum_{k=1}^N\frac1{(3n-1)(3n+1)}
&\le\sum_{k=1}^N\frac1{(3n-2)(3n+1)}\\
&=\frac13\sum_{k=1}^N\left(\frac1{3n-2}-\frac1{3n+1}\right)\\
&=\frac13\left(1-\frac1{3N+1}\right)
\end{align}
$$
Therefore,
$$
\sum_{k=1}^\infty\frac1{(3n-1)(3n+1)}\le\frac13
$$

Evaluating the Sum
Using $(9)$ from this answer,
$$
\begin{align}
\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-1}\right)
&=\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/3}\right)\\
&=\frac13H_{-1/3}\\[3pt]
&=-\frac12\log(3)+\frac\pi{6\sqrt3}\tag{1}
\end{align}
$$
Subtracting $(8)$ from $(7)$ in that same answer gives
$$
\begin{align}
\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n+1}\right)
&=\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-2}\right)+\overbrace{\sum_{n=1}^\infty\left(\frac1{3n-2}-\frac1{3n+1}\right)}^{\text{telescoping series}}\\
&=\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-2/3}\right)+1\\
&=\frac13H_{-2/3}+1\\[3pt]
&=-\frac12\log(3)-\frac\pi{6\sqrt3}+1\tag{2}
\end{align}
$$
Partial fractions and $(1)$ and $(2)$ yields
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+1)}
&=\frac12\sum_{n=1}^\infty\left(\frac1{3n-1}-\frac1{3n+1}\right)\\
&=\frac12\left[\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n+1}\right)
-\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-1}\right)\right]\\[3pt]
&=\frac12-\frac\pi{6\sqrt3}\tag{3}
\end{align}
$$
A: The method proposed is one way forward.  To carry out the integration, we use partial fraction expansion to write
$$\begin{align}
\int_1^{\infty}\frac{1}{(3x-1)(3x+1)}\,dx&=\int_1^{\infty}\left(\frac{1/2}{3x-1}-\frac{1/2}{3x+1}\right)\,dx\\\\
&=\left.\frac12\log\left(\frac{3x-1}{3x+1}\right)\right|_{1}^{\infty}\\\\
&=\frac12 \log 2
\end{align}$$
and conclude that the series converges.

At the request of the OP, we use limit comparison test to show convergence.  We recall that 
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
Since $$\lim_{n\to \infty}\frac{1/n^2}{1/(9n^2-1)}=9<\infty$$then the series of interest converges.
