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

An Integral Trick
Since
$$
\int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n
$$
for $n\gt0$, we can write
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+2)}
&=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\
&=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\
&=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\
&=\frac16
\end{align}
$$
A: Since $\int_{0}^{1}x^k\,dx = \frac{1}{k+1}$,
$$\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)=\frac{1}{3}\int_{0}^{1}x^{3n-2}(1-x^3)\,dx,$$
so, summing over $n$:
$$\sum_{n=1}^{+\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\int_{0}^{1}x\,dx=\frac{1}{6}.$$
A: 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 $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}$$$$=\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{3n-1}-\frac{1}{3n+2}\right) $$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\! \cdot \! ........ +\left(\frac{1}{3n-4}-\frac{1}{3n-1}\right)+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\frac{1}{2} -\frac{1}{3n+2}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2} -0\right]$$ $$=\color{blue}{\frac{1}{6}}$$
A: Actually writing it as an integral, as asked for:
$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{1}^{\infty} \frac{1}{(3\lfloor x\rfloor-1)(3\lfloor x\rfloor+2)} dx$$
This probably won't help with finding the value, though.
A: Yes you can use the Euler Maclaruin formula to write the sum as an integral plus an infinite number of derivatives. I remember deriving this for my self when I was younger and being very pleased with myself. 
A: 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 $x \to 1$
$$\sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)} = \log(4)$$
$$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)}$$
$$\psi(\frac{1}{2}) = -\gamma - \log(4)$$
A: 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 $f (k)$. 
