Series Convergence, Comparison theorem $\sum_{n=2}^{\infty} \frac{1}{n(n-1)}$ The series must converge by the comparison theorem, but I'm struggling to find a convergent series I could compare it to. 
Also, I wonder if integral test would work. 
Thanks for help!
 A: For any $n >1$ $$(n-1)^2\le n(n-1) \le n^2$$ This implies 
$$\sum_{n=2}^{\infty} \frac{1}{n^2}<\sum_{n=2}^{\infty} \frac{1}{n(n-1)}<\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}=\sum_{n=1}^{\infty} \frac{1}{n^2}$$ which finally makes 
$$\sum_{n=1}^{\infty} \frac{1}{n^2}-1<\sum_{n=2}^{\infty} \frac{1}{n(n-1)}<\sum_{n=1}^{\infty} \frac{1}{n^2}$$ and you know that $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac {\pi^2}6$.
A: The integral test would work actually, so for that you want to see if the integral of $\int_2 ^\infty \frac{1}{n(n-1)} dn$ converges.
After some simple calculations, you should get the answer to be $ln(2)$ (I will leave this as an exercise for you to verify).
So since the integral converges, that means the series converges as well.
A: This is another criterion, suppose we want to find a constant $c$ for which $$n(n-1) \ge cn^2, \ n=2,3,4, \cdots \tag{1}$$
Putting $n=2$ in (1), we obtain $$2 \ge 4c \tag{2}$$ Which implies that $$c \le 0.5 $$ We can choose $c=.5/2=0.25=\frac{1}{4}$. Hence, we have $$n(n-1)\ge \frac{n^2}{4}, \ n=2,3,4,\cdots \tag{3}$$ Transposing (3) we have $$\frac{1}{n(n-1)} \le \frac{4}{n^2}$$. Hence, you have got a series $\displaystyle \sum _{n=2} ^{\infty} \frac{4}{n^2}$, which converges by p-series test and combination theorem.
