let's take the sum:
$$\displaystyle \sum_{n=1}^{\infty}\frac{1}{9n^2 + 3n – 2}\\\implies 9n^2 + 3n – 2 = 9n^2 + 6n – 3n – 2 = 3n(3n + 2) – (3n + 2) = (3n – 1)(3n + 2)$$
The simplest way would be to use partial fractions, and then convert this into a telescoping series. Which makes the sum extremely simple, but I am looking for a way in, which I could use integrals, perhaps a definite integral in, which I could derive this sum. Any ideas?
The $f(n)$ is $f(n) = \frac{1}{(3n-1)(3n+2)}$
So can we use integrals?
Best Answer
Hint. You may write $$\begin{align} \sum_{n=1}^{\infty}\frac{1}{9n^2 + 3n - 2} &=\frac{1}{3}\sum_{n=1}^{\infty}\left(\frac{1}{3n - 1}-\frac{1}{3n + 2}\right)\\\\ &=\frac{1}{3}\sum_{n=1}^{\infty}\int_0^1\left(x^{3n-2}-x^{3n+1}\right){\rm d}x\\\\ &=\frac{1}{3}\int_0^1\sum_{n=1}^{\infty}\left(x^{3n-2}-x^{3n+1}\right){\rm d}x\\\\ &=\frac{1}{3}\int_0^1x\frac{1-x^3}{1-x^3}{\rm d}x\\\\ &=\frac{1}{3}\int_0^1x\:{\rm d}x\\\\ &=\frac{1}{6} \end{align} $$ The method by telescoping terms would be more direct here.