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I'm completely lost. Wolfram alpha says that partial sums look like this: $$\sum_{n=2}^{+\infty}\frac{1}{(n^2-1)}=\frac{3m^2-m-2}{4m(m+1)}$$ From here, the sum is clearly equal to $\frac34$. But how on earth does one come up with such formula? Maybe I'm missing something obvious here, but I can't figure it out. Thanks!

Dunno
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    Use partial fractions and you will as lab has said its a telescoping series. – user60887 Dec 05 '13 at 17:37
  • I think you can also get this from a fourier series of $\cos x$ or $\cosh x$ or something to that effect, but I can't recall. – abnry Dec 05 '13 at 17:43

3 Answers3

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HINT:

If $n^2\ne1\iff n\ne\pm1,$

$$\frac1{n^2-1}=\frac12\frac{(n+1)-(n-1)}{(n-1)(n+1)}=\frac12\left(\frac1{n-1}-\frac1{n+1}\right)$$

Can you recognize the Telescoping series ?

lab bhattacharjee
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  • Good hint...you beat me to it. –  Dec 05 '13 at 17:40
  • Interesting, we didn't have it on our lecture... Thanks for the link, it's very helpful! – Dunno Dec 05 '13 at 17:44
  • @Dunno, a few similar Questions : http://math.stackexchange.com/questions/569886/determine-whether-the-series-is-convergent-or-divergent-by-expressing-s-n-as-a http://math.stackexchange.com/questions/475256/if-this-is-a-telescoping-series-then-how-does-it-collapse-frac3r1rr-1r, http://math.stackexchange.com/questions/425966/finite-series-reciprocals-of-sines, http://math.stackexchange.com/questions/462082/how-to-solve-this-prod-infty-n-2-fracn3-1n31 http://math.stackexchange.com/questions/398075/value-of-series-partialsum – lab bhattacharjee Dec 05 '13 at 17:50
  • http://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2, http://math.stackexchange.com/questions/571973/evaluate-sum-limits-n-1-infty-fracnn4n21, http://math.stackexchange.com/questions/463308/how-to-find-the-sum-displaystyle-sumn-k-1-k2k1k – lab bhattacharjee Dec 05 '13 at 17:51
  • haha wow, thanks! It's always good to know I'm not the only one who doesn't see something. – Dunno Dec 05 '13 at 17:52
  • @Dunno, in general, most of us are mediocre in some sense. We shall find people superior to & inferior to us most of the times :) – lab bhattacharjee Dec 05 '13 at 18:12
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I'll give you a hint. Factor the bottom to get $$\frac{1}{(n+1)(n-1)}$$ Now use partial fractions to write $$\frac{1}{(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ Can you compute the $M$th partial sum rewriting in this way? Look for cancellations.

  • One small comment...just rememeber to calculate the limit as $M\to \infty$ when your done computing the $M$th partial sum. You will usually get a number $+$ something that goes to $0$, but it is possible to have a situation where you have a number $+$ a nonzero limit. –  Dec 05 '13 at 17:43
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The formula you obtained is for a summation from 2 to m and it is correct. If "m" goes to an infinite value, the limit is 3/4. Could you check what you submitted to WA ? Thanks for letting me know.

Claude Leibovici
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