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?
$\endgroup$ 92 Answers
$\begingroup$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
$$\lim_{N\rightarrow\infty}\sum_{n=3}^N\frac{1}{n(n-1)}$$
$\endgroup$ $\begingroup$Look at the partial sums.
Since $\frac{1}{n(n-1)} =\frac1{n-1}-\frac1{n}$, the sum of the first $M$ terms is
$\begin{align} \sum_{n = 3}^M \frac{1}{n(n-1)} &=\frac1{2*3}+\frac1{3*4}+\frac1{4*5}+...+\frac1{(M-1)*M}\\ &=(\frac1{2}-\frac1{3})+(\frac1{3}-\frac1{4})+(\frac1{4}-\frac1{5})+...+(\frac1{M-1}-\frac1{M})\\ &=\frac1{2}+(-\frac1{3}+\frac1{3})+(-\frac1{4}+\frac1{4})+... +(-\frac1{M-1}+\frac1{M-1})-\frac1{M} \quad\quad (*)\\ &=\frac1{2}-\frac1{M}\\ \end{align} $
The rearrangement in the line marked "$(*)$" is allowed because the sum has a finite number of terms, and finite sums can be rearranged in any way.
Note that all terms except the first $\frac1{2}$ and last $\frac1{M}$ are canceled out, so the final result for the partial sum of the first $M$ terms is $\sum_{n = 3}^M \frac{1}{n(n-1)} =\frac1{2}-\frac1{M} $.
The limit of this as $M \to \infty$ is $\frac1{2}$.
$\endgroup$ 1
