The Finite Sum $\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$ and failure to Telescope

Question

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

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

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

3. 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.

Answer 1

Here's another answer: it's known that $$1+2x^2\sum_{v=1}^{\infty}\frac1{x^2-v^2} = \pi x\cot(\pi x)$$ See here How did Euler prove the partial fraction expansion of the cotangent function: $\pi\cot(\pi z)=\frac1z+\sum_{k=1}^\infty(\frac1{z-k}+\frac1{z+k})$? for example. Setting $x=2/3$ gives $$1+2\sum_{v=1}^{\infty}\left(\frac1{3v+2}-\frac1{3v-2}\right)=\frac{2\pi}{3}\cot(\frac{2\pi}{3})=\frac{-2\pi\sqrt3}{9}$$, from which, after simple calculations:$$1/4\sum_{v=1}^{\infty}\left(\frac1{3v-2}-\frac1{3v+2}\right)=\frac{2\pi \sqrt3+9}{72}$$

Answer 2

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.

Answer 3

Regarding the infinite series: as you have already found, \begin{align*} \sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)} &= \frac{1}{4}\bigg(\sum_{r=1}^{n}\frac{1}{3r-2}-\Big(\sum_{r=0}^{n}\frac{1}{3r+2} - \frac12 \Big)\bigg) \\ &= \frac{1}{4}\bigg(\frac12+\sum_{n=1}^{\infty}\frac{\chi_{-3}(n)}n \bigg), \end{align*} where \begin{cases} 1, &\text{if } n\equiv1\pmod3, \\ -1, &\text{if } n\equiv2\pmod3, \\ 0, &\text{if } 3\mid n \end{cases} is the nonprincipal Dirichlet character of conductor $3$. The infinite series is a well-known constant $$ \sum_{n=1}^{\infty}\frac{\chi_{-3}(n)}n = L(1,\chi_{-3}) = \frac\pi{3\sqrt3}, $$ which recovers your formula (with a sign typo corrected).

Answer 4

Since the Digamma function satisfies the functional equation $$ \Delta \psi (z) = \psi (z + 1) - \psi (z) = {1 \over z} $$ Then its Anti-Delta is $$ \psi (z) = \Delta ^{\, - 1} {1 \over z} = \sum\nolimits_{\;z}^{} {{1 \over z}} $$

Using this concept, the sum is defined also for real (and in this case also complex) lower and upper bound, as $$ \sum\nolimits_{\;z = a}^{\;b} {{1 \over z}} = \psi (b) - \psi (a) $$

So your sum gets converted into the Anti-Delta concept as follows $$ \eqalign{ & {1 \over 4}\left( {\sum\limits_{r = 1}^n {{1 \over {3r - 2}}} - \sum\limits_{r = 1}^n {{1 \over {3r + 2}}} } \right) = \cr & = {1 \over 4}\left( {\sum\nolimits_{r = 1}^{n + 1} {{1 \over {3r - 2}}} - \sum\nolimits_{r = 1}^{n + 1} {{1 \over {3r + 2}}} } \right) = \cr & = {1 \over {12}}\left( {\sum\nolimits_{r = 1}^{n + 1} {{1 \over {r - 2/3}}} - \sum\nolimits_{r = 1}^{n + 1} {{1 \over {r + 2/3}}} } \right) = \cr & = {1 \over {12}}\left( {\sum\nolimits_{r = 1/3}^{n + 1/3} {{1 \over r}} - \sum\nolimits_{r = 5/3}^{n + 5/3} {{1 \over r}} } \right) = \cr & = {1 \over {12}}\left( {\psi (n + 1/3) - \psi (1/3) - \psi (n + 5/3) + \psi (5/3)} \right) \cr} $$ And in fact $$ \eqalign{ & \mathop {\lim }\limits_{n \to \infty } S = {1 \over {12}}\left( {\psi (5/3) - \psi (1/3)} \right) = \cr & = {1 \over {12}}\left( {\left( {{{\pi \sqrt 3 } \over 6} - {{3\ln \left( 3 \right)} \over 2} - \gamma + {3 \over 2}} \right) - \left( { - {{\pi \sqrt 3 } \over 6} - {{3\ln \left( 3 \right)} \over 2} - \gamma } \right)} \right) = \cr & = {1 \over {12}}\left( {{{\pi \sqrt 3 } \over 3} + {3 \over 2}} \right) \cr} $$

Answer 5

$$\sum_{r=1}^{\infty}\frac{1}{(3r-2)(3r+2)}=\sum_{r=1}^{\infty}\frac{1}{(9r^2-4)}=\frac{1}{9}\sum_{r=1}^{\infty}\frac{1}{r^2-(\frac{2}{3})^2}$$ then use the
$$\frac{1-\pi*x\cot\pi x}{2x^2}=\sum_{r=1}^{\infty}\frac{1}{r^2-x^2}$$ and put $x=\frac{2}{3}$

so the sum will be $$\frac{9+2\sqrt{3}\pi}{72}$$