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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 $\frac34$.

Regards,

Dennis

4 Answers 4

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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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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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    @BrianM.Scott: Yes. it gives just an upper bond. I noted here just for another approach if it is convergent. You are right absolutely.2012-11-02
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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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    @Dennis: You’re welcome!2012-11-02
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$\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} $