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.

We can indeed write the sum as an integral, after research. Consider:
Find: $\psi(1/2)$
By definition: 
$$\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$$
The required $z$ is $z = -\frac{1}{2}$
so let $z = -\frac{1}{2}$
$$\psi(1/2) = -\gamma + \sum_{n=1}^{\infty} \frac{-1}{2n(n - \frac{1}{2})}$$
Simplify this:
$$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{n(2n - 1)}$$
The sum seems difficult, but really isnt. 
We can telescope or:
$$\frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$
Let $x \rightarrow x^2$
$$\frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$
Integrate once:
$$\tanh^{-1}(x) = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}$$
Integrate again:
$$\sum_{n=1}^{\infty} \frac{x^{2n}}{(2n-1)(n)} = 2\int \tanh^{-1}(x) dx$$
From the tables, the integral of $\tanh^{-1}(x)$
$$\sum_{n=1}^{\infty} \frac{x^{2n}}{(2n-1)(n)} = \log(1 - x^2) + 2x\tanh^{-1}(x)$$
Take the limit as $x \to 1$
$$\sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)} = \log(4)$$
$$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)}$$
$$\psi(\frac{1}{2}) = -\gamma - \log(4)$$
