I am suppose to find $\sum_{n = 3}^\infty \frac{1}{n(n-1)}$
I am suppose to rewrite it as a telescoping series, but that isn't really defined so I don't know how to do that so I just copied the wikipedia page and get
$$\frac{-1}{n} + \frac{1}{n-1}$$
Ok whatever, I try and find the sum and i see both terms diverge. Is that the answer?
Best Answer
Notice that your sum can be written as
$$\sum_{n=3}^{\infty}\frac{1}{n(n-1)} = \sum_{n=3}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right).$$
Now, let's see what happens when we examine the first few terms in the series. If we add up the first three terms we have
$$\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right).$$
Notice that everything but the first and last terms cancel. What could you then conclude about the sum
$$\lim_{N\rightarrow\infty}\sum_{n=3}^N\frac{1}{n(n-1)}$$