0
$\begingroup$

I Can't see where the serie goes.

I opened N terms, but I'm having some issues. If someone can help. The serie is $$\sum_{n=2}^\infty\frac{2}{n^2-1}$$

$\endgroup$
  • $\begingroup$ Why don't u try with $$\dfrac2{n^2-1}=\dfrac{n+1-(n-1)}{(n+1)(n-1)}=f(n-1)-f(n+1)$$ where $$f(m)=\dfrac1m$$ $\endgroup$ – lab bhattacharjee Jun 7 '17 at 3:23
2
$\begingroup$

Hint: $$\frac{2}{n^2-1}=\frac{2}{(n-1)(n+1)}=\frac{1}{n-1}-\frac{1}{n+1}$$ Can you see how this is a telescopic sum?

Addendum: Looking at the $N$-th partial sum we have $$\begin{align}\sum_{n=2}^N\left(\frac{1}{n-1}-\frac{1}{n+1}\right)&=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+...+\\&\ \ \ \ \ \left(\frac{1}{N-3}-\frac{1}{N-1}\right)+\left(\frac{1}{N-2}-\frac{1}{N}\right)+\left(\frac{1}{N-1}-\frac{1}{N+1}\right)\\&=1+\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...\end{align}$$ Do you see which terms will cancel out (Hint: there should be two terms left over on each end of the sum)?

$\endgroup$
  • $\begingroup$ I did the partial fraction decomposition, but when i opened term by term i can't see the result $\endgroup$ – Hamilton Junior Jun 7 '17 at 3:27
  • $\begingroup$ @HamiltonJunior see my addendum $\endgroup$ – Dave Jun 7 '17 at 3:34
  • $\begingroup$ now i saw and two terms left if i correct. $ 1 $ and $ \frac{1}{2}$, Correct? $\endgroup$ – Hamilton Junior Jun 7 '17 at 3:37
  • $\begingroup$ Those are not quite the only terms left in the $N$-th partial sum. However, after taking the limit to infinity these are the remaining terms. Can you see what the other two terms remaining in the $N$-th partial sum should be? $\endgroup$ – Dave Jun 7 '17 at 3:38
  • $\begingroup$ $-\frac{1}{N}$ and $\frac{1}{N-3}$ ? $\endgroup$ – Hamilton Junior Jun 7 '17 at 3:42
1
$\begingroup$

$S_k=\sum_{n=2}^k\{\frac{1}{n-1}-\frac{1}{n+1}\}$

$=1+\frac{1}{2}-\frac{1}{k}-\frac{1}{k+1}$

Hence series converges to $\frac{3}{2}$

$\endgroup$
1
$\begingroup$

Hint: $$\int_0^1x^{n-2}-x^ndx=\frac{2}{n^2-1}$$ using this hint we will get $$\sum_{n=2}^{\infty}{\frac{2}{n^2-1}}=\int_0^1{\sum_{n=2}^{\infty}x^{n-2}-\sum_{n=2}^{\infty}x^ndx}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_0^1{\sum_{n=0}^{\infty}x^{n}-\sum_{n=2}^{\infty}x^ndx}\\ \ \ \ =\int_0^1{(1+x)dx}$$This is an integral one can do via standard calculus methods.

$\endgroup$

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.