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Show that this... $\sum_{k=2}^{2n+1}\frac{2}{k^2-1}= \frac{3}{2}-\frac{1}{2n+1}-\frac{1}{2n+2}$

I have already arrived at... 1+ $\frac{1}{2}$ + ($\frac{1}{2n-1}$ - $\frac{1}{2n+1}$) + ($\frac{1}{2n}$ - $\frac{1}{2n+2}$)

But I am unaware as to how to get the last part of the equation... $\frac{1}{2n+1}-\frac{1}{2n+2}$

L. Li
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1 Answers1

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HINT:

Note that

$$\frac2{k^2-1}=\frac{1}{k-1}-\frac1{k+1}=\left(\frac{1}{k-1}-\frac1k\right)+\left(\frac1k-\frac{1}{k+1}\right)$$

Now telescope two summations.

Mark Viola
  • 179,405
  • I have arrived at that point but I am not sure how 1/(2n) will be removed. – L. Li Feb 23 '18 at 02:09
  • Note that $\sum_{k=2}^{2n+1}\left(\frac1{k-1}-\frac1k\right)=1-\frac{1}{2n+1}$ and $\sum_{k=2}^{2n+1}\left(\frac1{k}-\frac1{k+1}\right)=\frac12-\frac{1}{2n+2}$. Adding gives $\frac32 -\frac{1}{2n+1}-\frac1{2n+2}$ as was to be shown! – Mark Viola Feb 28 '18 at 13:51
  • The last part is what I needed thank you! I think including your comment with the actual plug-ins of the number helped me see what the connection was between the two summations – L. Li Feb 28 '18 at 23:07
  • You're welcome. It was my pleasure to help! – Mark Viola Feb 28 '18 at 23:51