Why is $\sum_{n=1}^{k}\frac{1}{n^2+n}=\frac{k}{k+1}$ [duplicate]

Question

$$\sum_{n=1}^{k}\frac{1}{n^2+n}=\frac{k}{k+1}$$

I don't think that this summation requires too much context as this is a Q&A site, but I was just wondering why the summation is evaluated so nicely. I don't have too much knowledge and creativity in evaluating sums, so is there a bunch of equations that will make this make sense instantly? If so, I haven't found it on the internet.

$\frac{k}{k+1}$ looks a lot like the formula for the geometric series. I was wondering if $$\dfrac{1}{r}\left(\frac{1}{1-r}\right)=\frac{1}{-r^2+r}$$ had anything to do with this problem.

Answer 1

Hint

$$\frac{1}{n^2+n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$

Answer 2

Thank you to Surb for that hint. $$\begin{align*} \sum_{n=1}^{k}\frac{1}{n^2+n}&=\sum_{n=1}^{k}\frac{1}{n}-\frac{1}{n+1}\\ &=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+1} \\ &=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k-1}+\frac{1}{k}\\\\ &\phantom{\frac{1}{1}+}-\Big(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k-1}+\frac{1}{k}+\frac{1}{k+1}\Big)\end{align*}$$

Everything cancels out except for $\frac{1}{1}-\frac{1}{k+1}=\frac{k}{k+1}$.

Answer 3

The problem of this telescopic series was posed in Finnish Matriculation Exam, 2015 Spring, A-level mathematics, Problem set 15*, worth 9 points instead of ordinary 6. The problem set, question b) itself gives hints to the intuition behind it:

b.) Find $a, b \in \mathbb{R}$ s.t. $$ \frac{1}{n(n+1)} = \frac{a}{n}+\frac{b}{n+1}$$

The solution starts by expanding the terms:

$$ \frac{1}{n(n+1)} = {}^{n+1)} \frac{a}{n}+ {}^{n)} \frac{b}{n+1}$$ $$ \frac{1}{n(n+1)} = \frac{(a+b)n + a}{n(n+1)} $$

We have a linear system, we need to satisfy both $a+b = 0$ and $a=1$ which implies $b=-1$, the hint Surb gave.