1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Evaluate the limit of a series with an integral

  1. Aug 9, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n}\sqrt{\frac{4i}{n}}[/itex]

    3. The attempt at a solution

    This seems to give the right answer, 16/3, but I can't figure out why:

    [itex]\lim_{n \to \infty}\int_{1}^{n}\frac{4}{n}\sqrt{\frac{4x}{n}}dx[/itex]

  2. jcsd
  3. Aug 9, 2013 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I'd rather think that you have to interpret the sum as an approximation to the (Riemann) integral
    [tex]I=\int_0^{4} \mathrm{d} x \sqrt{x},[/tex]
    where the interval is divided in [itex]n[/itex] subintervals of equal size.
  4. Aug 9, 2013 #3


    User Avatar
    Science Advisor

    This is more than just approximating. If you have the function [itex]f(x)= \sqrt{x}[/itex], on the interval [0, 4], the Riemann sum, dividing [0, 4] into n equal subintervals, so that each subinterval has length 4/n and x= 4i/n, gives [tex]\sum{i= 1}^n \frac{4}{n}\sqrt{\frac{4i}{n}}[/tex]. As n goes to infinity, we are dividing the interval into more and more smaller and smaller intervals and the limit is the Riemann integral.
  5. Aug 9, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    vanhees71 said the sum approximates the Riemann integral; you're saying the limit of the sum is the Riemann integral. No contradiction there.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted