1. ## telescopic series

Hi guys, quick Question

For the series $\sum_{i=1}^\infty(\frac{1}{k}-\frac{1}{k+2})$ find a formula for $S_n$

I know that every number will cancel out but $1$ and $\frac{1}{2}$ but the answer is $1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}$.

How do we get $-\frac{1}{n+1}-\frac{1}{n+2}$?

Thx!

2. $S_n$ is not the sum of the infinite series, but the partial sum $\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+2})$.

You see it in the definition of the sum of an infinite series $\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+2})$ is defined as $\lim_{n \to \infty}S_n$, where $S_n$ is as above.

- Hollywood

3. Indeed. How do you get that partal sum? I understand the 1 and .5 part...

4. Everything cancels out except for 1 and 1/2 at the beginning and 1/(k+1) and 1/(k+2) at the end:

$S_n=\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+2})=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k+2}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=3}^{n+2}\frac{1}{k}$

$S_n=1+\frac{1}{2}+\sum_{k=3}^n\frac{1}{k}-\sum_{k=3}^n\frac{1}{k}-\frac{1}{n+1}-\frac{1}{n+2}$

$S_n=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}$

- Hollywood

5. I guess I'm still not sure how to get the right side. How do I produce it in the first place? I understand the 1 and .5. Thanks for your patience.

6. $\frac{1}{k}-\frac{1}{k+2}=\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2},$ there's two telescoping sums there.

7. Originally Posted by sfspitfire23
I guess I'm still not sure how to get the right side. How do I produce it in the first place? I understand the 1 and .5. Thanks for your patience.
If you look at the series this way:

$(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...$

the partial sums $S_n$ are:

$S_n=(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})$

It's no longer infinite, just the first n terms.

So you can see that the 1/3 in the first term cancels the one in the third term, the 1/4 in the second term cancels the one in the fourth term, and so on. But when we get to 1/(n+1), there is no term to cancel it, and the same for 1/(n+2).

Hope this helps - feel free to post again in this thread if you're still having trouble.

- Hollywood