Show convergence of a given series and find the limit.

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

Solution 1:

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 $[1,\infty]$.

Solution 2:

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 $n\to\infty$.

Solution 3:

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

Solution 4:

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

and use the telescoping property