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series: sum [n = 1, infty] [ (2n + 1) / (n^2(n + 1)^2 ]
- Thread starter mathhelp
- Start date
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 \(\displaystyle \L\\\frac{{\pi}^{2}}{6}\)
\(\displaystyle \L\\\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\)
The first half is a very popular series known as the Basel problem. The sum of the squares of the reciprocals is equal to \(\displaystyle \L\\\frac{{\pi}^{2}}{6}\)
Re: series
Hello, mathhelp!
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: \(\displaystyle \:S \:=\:1\)
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: \(\displaystyle \:S \:=\:1\)
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.
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.