I was wondering, for example,

Can:

∞∑n=11(3n−1)(3n+2)

Be written as an

Integral?To solve it. I amNOTtalking about a method for using tricks with integrals.But

actually writing an integral form.Like∞∑n=11(3n−1)(3n+2)=∫bag(x) dx

What are some

general tricksin finding infinite sum series.

**Answer**

**A General Trick**

A General Trick for summing this series is to use Telescoping Series:

∞∑n=11(3n−1)(3n+2)=13limN→∞N∑n=1(13n−1−13n+2)=13limN→∞[N∑n=113n−1−N∑n=113n+2]=13limN→∞[N−1∑n=013n+2−N∑n=113n+2]=13limN→∞[12−13N+2]=16

**An Integral Trick**

Since

∫∞0e−ntdt=1n

for n>0, we can write

∞∑n=11(3n−1)(3n+2)=∞∑n=113∫∞0(e−(3n−1)t−e−(3n+2)t)dt=13∫∞0e−2t−e−5t1−e−3tdt=13∫∞0e−2tdt=16

**Attribution***Source : Link , Question Author : Amad27 , Answer Author : robjohn*