Question

Prove
\[ \sum_{k=1}^\infty \frac 1 k -\frac 1{k+1} =1 \]

Use telescoping to show that \[ s_n =\sum_{k=1}^n \left ( \frac 1 k -\frac 1{k+1}\right )= 1 -\frac 1 {n+1}\]

Answer 1

@experimentX Try this one

Answer 2

yeah i am going for it ... though i would have to research that is telescopic method first.

Answer 3

This series is like 1-1/2+1/2-1/3+1/3+....So I guess there's a point

Answer 4

how do you write summation on latex??

Answer 5

\sum_{n=1}^\infty

Answer 6

summation for k=1 to n = 1/1 -1/2 + 1/2 -1/3 + 1/3 ...- 1/(n) + 1/(n) - 1/(1+n)

since all these appear on pairs, they cancel out each other ... except first and last term.
that gives Sn = 1 - 1/(n+1)

now if we put n = inf => Sn = 1 - 0 = 1

Answer 7

@experimentX you should say as \[ n\to \infty\]
\[s_n \to 1\]

Answer 8

yeah .... i think i understand.

Answer 9

\[Sn=\left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3} - \frac{1}{4} \right)+...+\left( \frac{1}{n} +\frac{1}{n+1}\right)\]
the terms from 1/2 until 1/n cancel each other, so we get
\[Sn = \left( 1 - \frac{1}{n+1} \right)=\frac{n}{n+1}\]\[\lim_{n \rightarrow \infty}Sn=\lim_{n \rightarrow \infty}\frac{n}{n+1}=1\]

Answer 10

similar question here
http://openstudy.com/study#/updates/4f7e8989e4b0bfe8930b4734