I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$s_n=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)=\left(\frac{1}{6}+\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{3}\right)+…+\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$
I then grouped the terms by their position in each partial sum,
First terms: $\frac{1}{6},\frac{1}{8},\frac{1}{10},\frac{1}{12}…$
Second terms: $\require\cancel{\cancel{\frac{1}{2}}},\cancel{\frac{1}{4}},\cancel{\frac{1}{6}},\cancel{\frac{1}{8}}…$
Third terms: $\cancel{-\frac{1}{2}},-\frac{1}{3},\cancel{-\frac{1}{4}},-\frac{1}{5}…$
Cancelling leaves the series: $$\sum_{n=3}^{\infty}\frac{1}{2n}-\sum_{n=1}^{\infty}\frac{1}{2n+1}$$
However I'm stuck here since I see a divergent harmonic series summed with another harmonic series but I know the original series in question is convergent to $\frac{1}{4}$. What can I do? I suspect my error can be fixed somehow by adjusting the bounds of the sums…? Thanks in advance.
Best Answer
Your partial fraction decomposition is fine, but you need not to break up into separate positive and negative sums. Instead, you need to cancel terms within the summation. $$\sum_{n=2}^{\infty}\bigg(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\bigg(\frac{1}{(n+1)}+\frac{1}{(n-1)}-\frac{2}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)-\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ Now notice that the negative part of the $n$ term of the sum cancels with the positive part of the $n+1$ term of the sum, so all the terms disappear except the positive part of the $n=2$ term. Our sum then equals $$\frac 12\left(\frac 11-\frac 12\right)=\frac 14$$