I've been working on this problem for hours and haven't seemed to get anywhere. I've split the summation in various ways, without really getting anywhere. Is there any simple method or identity that i'm missing here? Thank you for your help.

Calculate as a fraction


Use the shift from the proof of the Binomial Theorem.

  • $\begingroup$ My apologies,was just figuring out how to create summation notation on here! $\endgroup$ – Jake May 3 '17 at 1:19
  • $\begingroup$ Hint $\frac4{n^2-4}=\frac1{n-2}-\frac1{n+2}$. Summating from n=1 is wromg because must be $n\ne2$ $\endgroup$ – Minz May 3 '17 at 1:20
  • $\begingroup$ @Ross MIllikan is right Mathematica confirms $\endgroup$ – John Lou May 3 '17 at 1:23


$$\frac{1}{n^2-4}=\frac{1}{(n+2)(n-2)}=\frac{1}{4}\cdot \bigg(\frac{1}{n-2}-\frac{1}{n+2}\bigg)$$


The original answer was before the question update. But obviously, given now that the $n$ is specified to be able to take $2$, this summation is not defined.

  • $\begingroup$ This is probably what was intended, but the problem setter missed something. $\endgroup$ – Ross Millikan May 3 '17 at 1:21
  • $\begingroup$ @RossMillikan yea, this answer was posted before the question was updated. Previously, no index was specified, and thus, I would think the index should at least make the summation terms defined $\endgroup$ – Yujie Zha May 3 '17 at 1:22
  • $\begingroup$ Mathematica shows: $\sum\limits_{n=3}^{1000} {1 \over n^2 - 4} = {347248652417 \over 667999332000}$. $\endgroup$ – David G. Stork May 3 '17 at 1:30
  • $\begingroup$ @DavidG.Stork thanks for comments! Could you fix your MathJax formula? It is not rendering correctly, and I'm not sure if I get the intended formula.. $\endgroup$ – Yujie Zha May 3 '17 at 1:32
  • $\begingroup$ @Yujie Zha: Looks fine on my computer. What's the problem? $\endgroup$ – David G. Stork May 3 '17 at 1:33

There is no answer. The term with $n=2$ is undefined.

  • $\begingroup$ Yeah, that has confused me quite a bit. I'll mention that at n=2 it's undefined. However, I don't think it was the professor's intention, so I will probably try and solve the problem starting at n=3. $\endgroup$ – Jake May 3 '17 at 1:28
  • $\begingroup$ At $n=1$ it is $-\frac 13$, nicely defined. The problem is only at $n=2$. Yujie Zha's hint applies if you start at $n=3$ $\endgroup$ – Ross Millikan May 3 '17 at 1:30
  • $\begingroup$ Sorry was a typo. Thanks for your help! $\endgroup$ – Jake May 3 '17 at 1:31

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