<s>[INST] Checking for convergence of series To check the convergence of the series
$\displaystyle \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..\infty$
Attempt 1: Term $\displaystyle u_n= \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$
So the series becomes ...
$\displaystyle \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+..\infty$
This is an alternating series. So the series is said to be oscillatory or non-convergent.
Attempt 2: We have $\displaystyle u_n= \frac{1}{n^2}\frac{1}{1+1/n}$
and  $\displaystyle v_n= \frac{1}{n^2}$
Then $\displaystyle \lim_{n \to\infty}\frac{u_n}{v_n}=\lim_{n \to\infty}\frac{1}{1+1/n}=1$
Therefore, both $u_n$ and $v_n$ converge or diverge together. As $\displaystyle \sum v_n=\sum \frac{1}{n^2}$ converges, $\sum u_n$ must also converge.
Now my question is that which of this approaches is correct. The given answer says "convergent". So Attempt 2 should be correct. 
But, how do I know which method to choose for solving unknown problems. Please advise.
 [/INST] To check the convergence:
$$n(n+1)=n^2+n\geq n^2$$
then $$\frac{1}{n(n+1)}\leq \frac{1}{n^2}$$
and so $$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$
converge.
To calculate the limit,
$$\sum_{n=1}^N\frac{1}{n(n+1)}=\sum_{n=1}^N\frac{1}{n}-\sum_{n=1}^N\frac{1}{n+1}=\sum_{n=1}^N\frac{1}{n}-\sum_{n=2}^{N+1}\frac{1}{n}=1+\sum_{n=2}^N\frac{1}{n}-\sum_{n=2}^N\frac{1}{n}-\frac{1}{N+1}=1+\frac{1}{N+1}$$
When you make tighten $N\to\infty $, you finally obtain
$$\sum_{n=1}^\infty \frac{1}{n(n+1)}=1.$$
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