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.

• $$\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+1}=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=2}^{k+1}\frac{1}{n}=\frac1{?}-\frac1{??}$$ – Did Oct 2 '16 at 18:41
• This sort of series is known as telescoping series... – achille hui Oct 2 '16 at 18:42
• "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." Sorry but I find this odd. – Did Oct 2 '16 at 18:43
• You meant in terms of $k$ – zhw. Oct 2 '16 at 18:43

2 Answers

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, $$S=\lim_{n\to\infty}S_N=1$$

We can rewrite the sum

$$\sum_{k=1}^n\frac1{k(k+1)}=\sum_{k=0}^{n-1}\frac1{(k+1)(k+2)}=\sum\nolimits_0^nk^{\underline{-2}}\delta k=-k^{\underline{-1}}\Big|_0^n=1-\frac1{n+1}$$

where $k^{\underline{a}}$ represent a falling factorial.

Well, take this as an alternative to the telescoping sum. If you are curious about the mechanic you can read this.

• This is great! Thanks for the nice link. It's an awesome alternative to calculate the sum, and I'll definitely use this in other places! – Skeleton Bow Oct 2 '16 at 19:05