## anonymous 5 years ago summation series question: for the following equation do i use geometric series theorem, telescoping or harmonic methods? the problem is: (sorry idk how to illustrate summation LOL) infinity ----- `. { (1/2^k)-1/(2^k+1) } / ______ k=1

1. anonymous

Is your summation,$\sum_{k=1}^{\infty}\left( \frac{1}{2^k}-\frac{1}{2^{k+1}} \right)$?

2. anonymous

Because if it is, you should look at expanding some terms and seeing what happens:$(\frac{1}{2}-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{2^3})+(\frac{1}{2^3}-\frac{1}{2^4})$$=\frac{1}{2}+(-\frac{1}{2^2}+\frac{1}{2^2})+(-\frac{1}{2^3}+\frac{1}{2^3})-\frac{1}{2^4}$$=\frac{1}{2}-\frac{1}{2^4}$

3. anonymous

So for the first N summations, you would have,$\sum_{k=1}^{N}:=\frac{1}{2}-\frac{1}{2^{N+1}}$

4. anonymous

This is a partial sum, and the limit of the sequence of partial sums equals the limit of the series. So all you have to do is take N to infinity now.

5. anonymous

The series sums to 1/2.

6. anonymous

thanks! so i this would be geometric, a = 1/2 and r= 1/2 ?

7. anonymous

Technically, you should use mathematical induction to show for the N'th partial sum, but I don't think people are too precious about it.

8. anonymous

No, this is telescoping.

9. anonymous

alright, so the first few terms will cancell out and then i'll just use the initial term and subtract the function thing.. is there a way i can just look at the problem and know what it is? or am i going to have to use trial and error?

10. anonymous

Well, when I saw what you'd written, because of the minus sign between the two fractions, I thought to look at the first few terms to get a feel. You can then see a pattern of one term between the outer ones killing off another term. This is the definition of telescoping. You can't technically take the limit until you've set up a partial sum. Then you send N to infinity. For a geometric, you would see in the summand, or upon expanding a few terms, a common growth factor. You can test for geometric by dividing the (n+1)th term by the nth. If what you get is something independent of n, you have a common ratio, and your series is geometric.

11. anonymous

wonderful. thanks for your help! i'm still not looking forward to this exam tomorrow...

12. anonymous

Good luck :)