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.
A General Trick

A General Trick for summing this series is to use Telescoping Series:
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+2)}
&amp;=\frac13\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{3n-1}-\frac1{3n+2}\right)\\
&amp;=\frac13\lim_{N\to\infty}\left[\sum_{n=1}^N\frac1{3n-1}-\sum_{n=1}^N\frac1{3n+2}\right]\\
&amp;=\frac13\lim_{N\to\infty}\left[\sum_{n=0}^{N-1}\frac1{3n+2}-\sum_{n=1}^N\frac1{3n+2}\right]\\
&amp;=\frac13\lim_{N\to\infty}\left[\frac12-\frac1{3N+2}\right]\\
&amp;=\frac16
\end{align}
$$



An Integral Trick

Since
$$
\int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n
$$
for $n\gt0$, we can write
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+2)}
&amp;=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\
&amp;=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\
&amp;=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\
&amp;=\frac16
\end{align}
$$
This particular sum could be solved because you had two terms $ax+b$ and $ax+c$ and the difference between c and b is equal to a (I think it would work in a slightly more complicated way if it was a not-too-large multiple of a). 

If you want numerical values in general cases, and the sum doesn't converge quickly for your taste, or you want just a partial sum, you can use that 

$$\displaystyle f (k) = \int_{k-1/2}^{k+1/2} f(k) dx ≈ \int_{k-1/2}^{k+1/2} f(x) dx$$

and therefore

$$\displaystyle \sum_{k=n}^{m} f(k) ≈ \int_{n-1/2}^{m+1/2} f(x) dx$$

Assuming that you can solve the integral in closed form, if you let

$$\displaystyle g (k) = f(k) - \int_{k-1/2}^{k+1/2} f(x) dx$$

then 

$$\displaystyle \sum_{k=n}^{m} f(k) = \int_{n-1/2}^{m+1/2} f(x) dx + \sum_{k=n}^{m} g(k)$$

$g (k)$ will usually converge much faster than $f (k)$.