Calculate summation possible formula of geometric sum

Question

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, Oct 12 '16 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. — Ross Millikan, Oct 12 '16 at 15:00
See also: What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ or Partial sum of the series $\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$. — Martin Sleziak, Dec 24 '19 at 09:17

score 1 · Answer 1 · answered Oct 12 '16 at 14:58

1

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

answered Oct 12 '16 at 14:58

E.H.E

23,590

score 0 · Answer 2 · 2016-10-12T15:01:07.400

0

Interestingly, the formula does not simplify when you reduce to the common denominator:

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

doesn't look easier.

By if you expand a few terms using the original form, the solution all of a sudden looks trivial.

edited Oct 12 '16 at 15:01

answered Oct 12 '16 at 14:56

Calculate summation possible formula of geometric sum

2 Answers2