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.

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 $f (k)$. 
