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Given the Series $$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$

How exactly would I find out the limit is $\frac34$ as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to $\frac34$.



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up vote 1 down vote accepted

Partial fractions:

$$\sum_{k=1}^n \frac{1}{k(k+2)}=\frac12\sum_{k=1}^n\left(\frac1k-\frac1{k+2}\right)\;.$$

Now telescope, and take the limit as $n\to\infty$.

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Thanks, I haven't heard about telescopic sums before, but from what I understand I should simply calculate $$\frac11 - \frac{1}{n+2}$$ where N is infinity, yes? – Dennis Röttger Nov 2 '12 at 18:26
@Dennis: Not quite: $$\sum_{k=1}^n\left(\frac1k-\frac1{k+2}\right)=1+\frac12-\frac1{n+1}-\frac1{n+2}‌​\;,$$ with two uncancelled terms at each end. The last two tend to $0$ as $n\to\infty$, so the limit is $1+\frac12=\frac32$, and half of that is indeed $\frac34$. – Brian M. Scott Nov 2 '12 at 18:29
Oh, you took the positive terms of the first 2 members and took the last 2 negative ones, that does make sense obviously. Thanks a lot! – Dennis Röttger Nov 2 '12 at 18:32
@Dennis: You’re welcome! – Brian M. Scott Nov 2 '12 at 18:33

Hint: Use the integral test for verifying the series is convergent. Take $f(x)=\frac{1}{x(x+2)}$. $f(x)$ is positive and monotonic decreasing on $[1,\infty]$.

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That won’t give the actual limit, though. – Brian M. Scott Nov 2 '12 at 18:20
@BrianM.Scott: Yes. it gives just an upper bond. I noted here just for another approach if it is convergent. You are right absolutely. – Babak S. Nov 2 '12 at 18:26

$$\frac2{k(k+2)}=\frac1k-\frac1{k+2}\implies2\sum_{k=1}^n\frac1{k(k+2)}=1+\frac12-\frac1{n+1}-\frac1{n+2} $$

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Hint: rewrite $$ \frac{1}{k(k+2)}=\frac{1}{2k}-\frac{1}{2(k+2)} $$

and use the telescoping property

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