TITLE: Show convergence of a given series and find the limit.
QUESTION [2 upvotes]: 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

REPLY [2 votes]: 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 $n\to\infty$.

REPLY [2 votes]: 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 $[1,\infty]$.

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

REPLY [1 votes]: Hint: rewrite $$
\frac{1}{k(k+2)}=\frac{1}{2k}-\frac{1}{2(k+2)}
$$
and use the telescoping property
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