A community for students.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is your summation,\[\sum_{k=1}^{\infty}\left( \frac{1}{2^k}\frac{1}{2^{k+1}} \right)\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because if it is, you should look at expanding some terms and seeing what happens:\[(\frac{1}{2}\frac{1}{2^2})+(\frac{1}{2^2}\frac{1}{2^3})+(\frac{1}{2^3}\frac{1}{2^4})\]\[=\frac{1}{2}+(\frac{1}{2^2}+\frac{1}{2^2})+(\frac{1}{2^3}+\frac{1}{2^3})\frac{1}{2^4}\]\[=\frac{1}{2}\frac{1}{2^4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So for the first N summations, you would have,\[\sum_{k=1}^{N}:=\frac{1}{2}\frac{1}{2^{N+1}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The series sums to 1/2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks! so i this would be geometric, a = 1/2 and r= 1/2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Technically, you should use mathematical induction to show for the N'th partial sum, but I don't think people are too precious about it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, this is telescoping.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wonderful. thanks for your help! i'm still not looking forward to this exam tomorrow...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.