A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
Use identity \(\dfrac{1}{k(k+1) }= \dfrac{1}{k}\dfrac{1}{k+1}\) to calculate \(\sum_{k=1}^n \dfrac{1}{k(k+1)}\)
Please, help
Loser66
 one year ago
Use identity \(\dfrac{1}{k(k+1) }= \dfrac{1}{k}\dfrac{1}{k+1}\) to calculate \(\sum_{k=1}^n \dfrac{1}{k(k+1)}\) Please, help

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It's easy to see that it is a telescoping series. If I let \(a_k = 1/k\) then \(a_{k+1}= 1/(k+1)\). What is wrong with it?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hence the sum is = \(a_n  a_0 = (1/n) 1 = \dfrac{1n}{n}\) but it is wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that sum supposed to be from \(k=1\) to \(n\)?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0the series is like  11/2 + 1/2 1/3.......1/n 1/(n+1) =11/(n+1) =(n)/(n+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider: $$\sum_{k=1}^n\left(\frac1k\frac1{k+1}\right)=1\frac12+\frac12\frac13+\dots+\frac1n\frac1{n+1}=1\frac1{n+1}=\frac{n}{n+1}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@imqwerty @oldrin.bataku Yes, it is, if we expand it. My question is why cannot I apply formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, you incorrectly collapsed the telescoping series. it actually collapses to: $$\sum_{k=1}^n\frac1k\sum_{k=1}^n\frac1{k+1}=\left(1+\sum_{k=2}^n\frac1k\right)\left(\sum_{k=2}^n\frac1n+\frac1{n+1}\right)=1\frac1{n+1}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what you mean by 'formula', there's no general formula for a telescoping series in that sense  telescoping just means it can be rewritten as a series in which interior terms cancel out. exactly what gets left over and in what order does not really follow from a 'formula', it's dependent on the form of the series

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, that should be \(\sum_{k=2}^n\frac1k\) both times there, not \(\sum\frac1n\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I got what is wrong with it. I am sorry for my silly. Yes, we have \(\sum_{k =1}^n (a_k  a_{k1})= a_n a_0\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0but it is \(a_k a_{k1}\) that is the term  its previous term while I interpreted as term  next term. I am sorry.
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.