Is it possible to write a sum as an integral to solve it? I was wondering, for example,
Can:
$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$
Be written as an Integral? To solve it.  I am NOT talking about a method for using tricks with integrals.
But actually writing an integral form. Like
$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{a}^{b} g(x) \space dx$$
What are some general tricks in finding infinite sum series.

In such cases, the partial fractions of general term (i.e. $n^{th}$ term ) of the infinite-series are very useful.
Given that $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\sum_{n=1}^{\infty} T_{n}$$
Where, $T_{n}$ is the   $n^{th}$ term of the given series which can be easily expressed in the partial fractions as follows $$T_{n}=\frac{1}{(3n-1)(3n+2)}$$$$=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)$$  Now, we have $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}$$$$=\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{3n-1}-\frac{1}{3n+2}\right) $$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\! \cdot \! ........ +\left(\frac{1}{3n-4}-\frac{1}{3n-1}\right)+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\frac{1}{2} -\frac{1}{3n+2}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2} -0\right]$$ $$=\color{blue}{\frac{1}{6}}$$
