Is it possible to write a sum as an integral to solve it?

A General Trick

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

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An Integral Trick

Since $$ \int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n $$ for $n\gt0$, we can write $$ \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+2)} &=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\ &=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\ &=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\ &=\frac16 \end{align} $$

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

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.