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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

    Mentallic

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    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

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    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?##

    RGV
     
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