How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
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1$\begingroup$ Aside: sums like this are often rather nontrivial to compute in familiar closed forms; e.g. see the sum of $1/n^2$. Those you can compute (like this one) in a simple fashion rely on being in the right form to use various tricks. $\endgroup$– user14972Aug 14, 2015 at 8:08
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3 Answers
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$$\frac{5}{9}\sum \frac{1}{4x^2-1}$$ $$=\frac{5}{18}\sum \frac{1}{2x-1}-\frac{1}{2x+1}$$ $$=\frac{5}{18}[ \frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}...]$$ $$=\frac{5}{18}[\frac{1}{5}]$$ $$=\frac{1}{18}$$
answered Aug 14, 2015 at 4:03
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Hint: See if the series telescopes if you were to approach via partial fraction decomposition.
answered Aug 14, 2015 at 4:03
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Notice, we have $$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}$$ $$=\frac{5}{9}\sum_{n=3}^{\infty}\frac{1}{4n^2-1}$$
$$=\frac{5}{9}\sum_{n=3}^{\infty}T_n$$ $$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\lim_{n\to \infty}\sum_{n=3}^{n}T_n\tag 1$$
Where, $T_n$ is nth term of the series
given as
$$T_n=\frac{1}{4n^2-1}= \frac{1}{(2n-1)(2n+1)}$$ $$=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$ Now, setting $n=3, 4, 5, \dots n$, we get $$T_1=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)$$ $$T_2=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{9}\right)$$ $$T_3=\frac{1}{2}\left(\frac{1}{9}-\frac{1}{11}\right)$$ $$..................$$
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$$T_{n-1}=\frac{1}{2}\left(\frac{1}{2n-3}-\frac{1}{2n-1}\right)$$
$$T_{n}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
Now, adding all the terms column-wise we get sum of all $(n-2)$ terms as follows $$\sum_{n=3}^{n} T_n=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$
Now, taking limit as $n\to \infty$, we get $$\lim_{n\to \infty}\sum_{n=3}^{n} T_n=\lim_{n\to \infty}\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\lim_{n\to \infty}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\left(\frac{1}{5}-0\right)=\frac{1}{10}$$ Now, setting the above value in (1), we get
$$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\times \frac{1}{10}$$ $$=\frac{1}{18}$$
answered Aug 14, 2015 at 7:19
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$\begingroup$ Why is $\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$? $\endgroup$– IvoAug 14, 2015 at 8:48
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$\begingroup$ you change $$\frac{1}{(2n-1)(2n+1)}$$ into partial fractions which is $$\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n-1}\right)$$ You can also verify the equality $\endgroup$ Aug 14, 2015 at 8:53
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$\begingroup$ I think I get the telescoping series part, but would you mind showing the partial fraction part. I don't know why, but I can't seem to solve for A and B (I know this is trivial, I'm just a bit drained). $\endgroup$– PoGaMiAug 15, 2015 at 2:03
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$\begingroup$ Never mind, I worked it out after a short break. Thank you for the solution! $\endgroup$– PoGaMiAug 15, 2015 at 3:12