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.

Solution 1:

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 $$ \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+2)} &=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\ &=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\ &=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\ &=\frac16 \end{align} $$

Solution 2:

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 $n$: $$\sum_{n=1}^{+\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\int_{0}^{1}x\,dx=\frac{1}{6}.$$

Solution 3:

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.

Solution 4:

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 $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{3n-1}-\frac{1}{3n+2}\right) $$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\! \cdot \! ........ +\left(\frac{1}{3n-4}-\frac{1}{3n-1}\right)+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\frac{1}{2} -\frac{1}{3n+2}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2} -\frac{1}{\infty}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2}\right]=\color{blue}{\frac{1}{6}}$$