1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of a sum , what am I doing wrong?

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    ## lim_{n \rightarrow \infty}{\frac{1}{n^2} \sum_{k=1}^{n} ke^{\frac{k}{n}}} ##

    2. Relevant equations


    3. The attempt at a solution
    ## lim_{n \rightarrow \infty}{\frac{1}{n^2} \sum_{k=1}^{n} ke^{\frac{k}{n}}} \\
    = lim_{n \rightarrow \infty}{\frac{1}{n^2} (1e^{\frac{1}{n}}+2e^{\frac{2}{n}}+3e^{\frac{3}{n}}+\ldots+ne^{\frac{n}{n}})} \\
    = lim_{n \rightarrow \infty}{\frac{e^{\frac{1}{n}}}{n^2} +\frac{2e^{\frac{2}{n}}}{n^2} + \ldots + \frac{e}{n} } = 0 ##

    I know the limit equals to 1 (Wolfram). Isn't the limit of a sum equals to the sum of the limits? What am I doing wrong?
     
    Last edited: Feb 2, 2016
  2. jcsd
  3. Feb 2, 2016 #2

    RUber

    User Avatar
    Homework Helper

    It looks like you are saying that since each individual term in the sum goes to zero, the infinite sum of those small terms will also be zero. This is not true.
     
  4. Feb 2, 2016 #3
    Thank you for the reply. It's a sum of ## n ## terms.

    Any tips on how to calculate an exact form of the sum: ## \sum_{k=1}^{n} ke^{\frac{k}{n}} ## ?
     
  5. Feb 2, 2016 #4

    RUber

    User Avatar
    Homework Helper

    I would recommend starting with the assumption that the exponential term is on the order of 1.
    What is
    ## \sum_{k=1}^n k ##?
    That should get you started in the right direction.
     
  6. Feb 2, 2016 #5
    The limit is ##\int_0^1 xe^x \ dx##. Check Rieman sums on Wikipedia to see how it works
     
  7. Feb 2, 2016 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For fixed ##n## this is just the sum ##\sum_{k=1}^n k x^k##, where we substitute ##x = e^{1/n}## at the end. That is a standard sum, and can using Google, for example. However, if you want to get it for yourself you can do it using a bit of calculus. For any ##x## (before the substitution) we have
    [tex] \sum_{k=1}^n k x^k = \sum_k x \frac{d\, x^k}{dx} = x \frac{d}{dx} \sum_{k=1}^n x^k.[/tex]
    Use the well-known formula for the sum ##\sum_{k=1}^n x^k##, take the derivative of that formula wrt ##x##, then multiply by ##x##. Finally, substitute ##x = e^{1/n}##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit of a sum , what am I doing wrong?
  1. What am I doing wrong? (Replies: 5)

  2. What am i doing wrong? (Replies: 2)

Loading...