Finding $\sum\limits_{n=1}^{k}\frac{1}{n(n+1)}$ in terms of $k$ So I found a question which asked me to find the sum
$$\sum_{n=1}^{k}\frac{1}{n(n+1)}$$
The only hint given was to rewrite the summation (the fraction after the sigma, does anyone know what it's called?) using fraction decomposition, so I did:
$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
So it becomes
$$\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
I then tried to find a pattern by writing the terms out beginning with $n=1$ but couldn't find anything that would help me find the sum.

You have completed the problem. The sum becomes
$$S=1-\frac12+\frac12-\frac13+\frac13-\frac14+\frac14+-\cdots$$
The $N^{th}$ partial sum is
$$S_N=1-\frac{1}{N+1}$$
Thus,
$$S=\lim_{n\to\infty}S_N=1$$
