The Finite Sum $\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$ and failure to Telescope I'm interested in when the partial fraction method of trying to get a series to telescope fails, and have alighted upon the interesting example of,
$$\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$$
The standard method of tackling this would be to use partial fractions to get,
$$\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}=\frac{1}{4}\big(\sum_{r=1}^{n}\frac{1}{3r-2}-\sum_{r=1}^{n}\frac{1}{3r+2}\big)$$
and then look at the fractions generated by this alternative representation of the sum,
$$=\frac{1}{4}\big(\big(\frac{1}{1}-\frac{1}{5}\big)+\big(\frac{1}{4}-\frac{1}{8}\big)+\big(\frac{1}{7}-\frac{1}{8}\big)+\big(\frac{1}{10}-\frac{1}{14}\big)+...+\big(\frac{1}{3n-2}-\frac{1}{3n+2}\big)\big)$$
To me, this does not obviously telescope.
Questions

*

*I am was wondering if anyone can see a way to get this to telescope.


*Failing that, have a nice explanation of why it will not telescope.


*Is there a way of evaluating the sum to $n$ terms via a different approach ?
An Observation
In exploring the series I put it into Wolfram Alpha which reported that the series is convergent and that,
$$\sum_{r=1}^{\infty}\frac{1}{(3r-2)(3r+2)}=\frac{1}{72}\big(2\sqrt{3}\pi+9\big)$$
Given the famous Basel problem for the sum of reciprocals of squares, the appearance of $\pi$ is not, perhaps, a surprise.

We can write the sum as
\begin{align*}
\sum_{r = 1}^{n}\dfrac{1}{(3r-2)(3r+2)} &= \dfrac{1}{4}\sum_{r = 1}^{n}\left[\dfrac{1}{(3r-2)} - \dfrac{1}{(3r+2)}\right]
\\
&= \dfrac{1}{4}\sum_{r = 1}^{n}\int_{0}^{1}(x^{3r-3}-x^{3r+1})\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\sum_{r = 1}^{n}(x^{3r-3}-x^{3r+1})\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\dfrac{(1-x^4)-(x^{3n}-x^{3n+4})}{1-x^3}\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\dfrac{1-x^4}{1-x^3}\,dx - \dfrac{1}{4}\int_{0}^{1}x^{3n}\dfrac{1-x^4}{1-x^3}\,dx.
\end{align*}
You can easily  show that $$0 \le \int_{0}^{1}x^{3n}\dfrac{1-x^4}{1-x^3}\,dx \le \int_{0}^{1}x^{3n}\sup_{x \in [0,1]}\left[\dfrac{1-x^4}{1-x^3}\right]\,dx = \int_{0}^{1}\dfrac{4}{3}x^{3n}\,dx = \dfrac{4}{3(3n+1)}$$ for all integers $n \ge 0$, and thus, $$\int_{0}^{1}x^{3n}\dfrac{1-x^4}{1-x^3}\,dx \to 0 \quad \text{as} \quad n \to \infty.$$
Therefore, $$\sum_{r = 1}^{\infty}\dfrac{1}{(3r-2)(3r+2)} = \dfrac{1}{4}\int_{0}^{1}\dfrac{1-x^4}{1-x^3}\,dx = \dfrac{9+2\pi\sqrt{3}}{72},$$ which can be easily evaluated using partial fractions.
