I was wondering, for example,
Can:
∞∑n=11(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
∞∑n=11(3n−1)(3n+2)=∫bag(x) dx
What are some general tricks in 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