series: sum [n = 1, infty] [ (2n + 1) / (n^2(n + 1)^2 ]

[mathhelp]

Find whether the following series converges or diverges. If the series converges, find its exact sum. Please show all work.

 

[galactus]

If you expand out your sum, you have:

\(\displaystyle \L\\\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\)

The first half is a very popular series known as the Basel problem. The sum of the squares of the reciprocals is equal to \(\displaystyle \L\\\frac{{\pi}^{2}}{6}\)

 

[soroban]

Re: series

Hello, mathhelp!

Find whether the following series converges or diverges.
If the series converges, find its exact sum.

\(\displaystyle \L S \;= \;\sum^{\infty}_{n=1}\,\frac{2n\,+\,1}{n^2(n\,+\,1)^2}\)

Applying Partial Fractions, we find that: $\displaystyle \:\frac{2n\,+\,1}{n^2(n\,+\,1)^2} \:=\:\frac{1}{n^2}\,-\,\frac{1}{(n\,+\,1)^2}$

Hence: $\displaystyle S \;= \;\left(\frac{1}{1^2}\,-\,\frac{1}{2^2}\right)\,+\,\left(\frac{1}{2^2}\,-\,\frac{1}{3^2}\right)\,+\,\left(\frac{1}{3^2}\,-\,\frac{1}{4^2}\right)\,+\,\left(\frac{1}{4^2}\,-\,\frac{1}{5^2}\right)\,+\,\cdots$

After the first term, all terms cancel out.

Therefore, the series converges to: $\displaystyle \:S \:=\:1$

 

[galactus]

I was going for the more complicated \(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{6}\) and \(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{(n+1)^{2}}=\frac{{\pi}^{2}}{6}-1\)

Therefore, \(\displaystyle \L\\\frac{{\pi}^{2}}{6}-(\frac{{\pi}^{2}}{6}-1)=1\)

If one makes the observation to the relationship with Euler's Basel series, then it is rather easy.