# Calculate summation possible formula of geometric sum

I need to calculate this:

$\sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})$

The answear is:

$1 - \frac{1}{n+1}$

But I have no idea how to get to that answear. I tried to simplify the equation to: $\frac{1}{kx^n+k}$ But now I don't know If I should put it inside the formula for geometric sums or if it's even possible.

Thank you!

• This is not geometric! Try writing it out for $n=2$. Then $n=3$. I think you'll see a pattern.
– lulu
Commented Oct 12, 2016 at 14:53
• This is a telescoping series where the second half of one term cancels with the first half of the next term. That leaves you with only the first half of the first term, the $1$, and the second half of the last term, the $-\frac 1{n+1}$, as they have nothing to cancel with. Commented Oct 12, 2016 at 15:00
• Commented Dec 24, 2019 at 9:17

$$\sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})=\sum_{k=1}^{n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k+1}=\sum_{k=0}^{n-1}\frac{1}{k+1}-\sum_{k=1}^{n}\frac{1}{k+1}$$ $$=1+\sum_{k=1}^{n-1}\frac{1}{k+1}-\sum_{k=1}^{n}\frac{1}{k+1}$$
$$\sum_{k=1}^n\frac1{k(k+1)}$$