$$\sum_{n=2}^\infty \frac{2}{n^2 - 1}.$$ I Tried setting it up as a telescoping sum as $$\frac{2}{n} - \frac{2}{n-1}.$$ but now i'm sure that cannot be correct. Mayhaps I need to complete the square? or pull out an N? I'm sorry for asking so many questions. I just really need to get a grasp on these sort of things...
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Also how do I format the question correctly? i can't make heads or tails of the help section. – Tibbiticus Oct 20 '13 at 23:37
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Are you familiar with partial fraction decomposition and Telescoping Sums? – imranfat Oct 20 '13 at 23:38
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I am with telescoping sums, i just don't know how to properly decompose it. i'm trying to format the question right now to look better – Tibbiticus Oct 20 '13 at 23:39
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haha! well if i can master the formatting on here surely i can master the problem! – Tibbiticus Oct 20 '13 at 23:41
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We know $$ \frac{2}{n^2-1} = \frac{2}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1} $$ for some $A,B$. Solving for $A,B$ we find $A=1,B=-1$. Thus we want to calculate $$ \sum_{n=2}^\infty (\frac{1}{n-1} - \frac{1}{n+1}) $$ which telescopes. All the terms cancel except for $\frac{1}{2-1}$ and $\frac{1}{3-1}$ so the sum is $\frac{3}{2}$.
nullUser
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OF COURSE ARG Factoring always gets me. All that no child left behind BS made my Algebra Skills Weak Thank you Very much! – Tibbiticus Oct 20 '13 at 23:44
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The sum is equal to $$\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+\cdots.$$ Observe the cancellations.
André Nicolas
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For every $m \ge 2$ we have \begin{eqnarray} S_m:&=&\sum_{n=2}^m\frac{2}{n^2-1}=\sum_{n=2}^m\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\\ &=&\sum_{n=1}^{m-1}\frac{1}{n}-\sum_{n=3}^{m+1}\frac{1}{n}=1+\frac12+\frac{1}{m}+\frac{1}{m+1}\\ &=&\frac32+\frac{1}{m}+\frac{1}{m+1}. \end{eqnarray} Taking the limit we get $$ \sum_{n=2}^\infty\frac{2}{n^2-1}=\lim_{m \to \infty}S_m=\frac32. $$
HorizonsMaths
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I was trying to solve this problem myself and I can get the 3/2 but I was wondering where the 2 on the top goes? Shouldn't the end result of the telescoping be multiplied by the 2? – Joshhw Jul 02 '14 at 19:33