5
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ MathJax hint: instead of using \bigg( and \bigg) for large parentheses, use \left( and \right). Then the parentheses are sized according to what is inside. If you make an edit so there are no more fractions inside the parentheses reduce in size appropriately. It works for paired things. $\endgroup$ – Ross Millikan Jul 29 '18 at 0:33
  • 1
    $\begingroup$ Thank you for the tip! @RossMillikan $\endgroup$ – coreyman317 Jul 29 '18 at 0:38
8
$\begingroup$

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 $$\frac 12\left(\frac 11-\frac 12\right)=\frac 14$$

$\endgroup$
2
$\begingroup$

$$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$

$$=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{2n}-\frac{1}{2n}\right)$$

$$=1/2\sum_{n=2}^{\infty}\left(\frac{1}{(n+1)}-\frac{1}{n}+\frac{1}{(n-1)}-\frac{1}{n}\right)$$

Do you see two telescoping series?

$\endgroup$
2
$\begingroup$

Alt. hint:   it is not necessary to do a full partial fractions decomposition, it's enough to telescope:

$$ \frac{1}{n^3-n} = \frac{1}{2}\frac{(n+1) - (n-1)}{(n-1)n(n+1)} = \frac{1}{2}\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.