1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Proof of an inequality involving a series (probably by induction)

  1. Sep 11, 2012 #1
    [tex]u_{n} = \sum_{k=1}^{n}\frac{1}{n+\sqrt{k}}[/tex]
    Proof that:
    [tex]\frac{n}{n+\sqrt{n}} \leq u_{n} \leq \frac{n}{n+1} [/tex]

    Ok, I've been working on that problem for about two hours now and I still don't have a clue how to proof this inequality.
    I guess it should be done by induction, but I have problems with the series, because I don't know how I could possibly pass from n to n+1, since the variable n is on the denominator.
    Perhaps there is a pretty easy solution to this problem, but any help would be welcome!
    (I'm sorry that I don't post my attempts at a solution, but I have to much of them and I don't believe that there is anything really productive

    Thanks in advance :)
  2. jcsd
  3. Sep 11, 2012 #2


    User Avatar
    Homework Helper

    What is [tex]\sum_{k=1}^{n}\frac{1}{n+1}[/tex] and how can you be sure it is greater than un?
  4. Sep 11, 2012 #3
    [tex]\sum_{k=1}^{n}\frac{1}{n+1} = \frac{n}{n+1}[/tex]
    [tex]n+\sqrt{k} \geq n+1[/tex] for every [itex]k\geq 1[/itex]
    Thus [tex]\frac{1}{n+\sqrt{k}} \leq \frac{1}{n+1}[/tex]
    and [tex]\sum_{k=1}^{n}\frac{1}{n+\sqrt{k}}\leq \sum_{k=1}^{n}\frac{1}{n+1}[/tex]

    And nearly the same works for the other part of the inequality

    So I guess I was totally mistaken with tryng to apply induction...
    Well thanks anyway! :)
  5. Sep 11, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The word you want is "prove", not "proof". Anyway, if $$t_k =\frac{1}{n + \sqrt{k}},$$ can you find a simpler quantity ##u_k## that bounds ##t_k## from above and is easy to sum? That is, can you think of a bound ##t_k \leq u_k , ## where ##u_k## is easier to deal with? Can you do something similar for a lower bound ##l_k \leq t_k?##

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook