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Proving an inequality, Real Analysis

  1. Sep 19, 2009 #1
    2. Relevant equations

    Prove the following for n > 1, n is a natural number.

    Sum (from i=1 to n) of:
    1/sqrt(i)

    is > then

    sqrt(n)



    3. The attempt at a solution

    To be honest I have spent 2+ hours on this with little results. I have tried induction on n, I tried showing one increases more every step then the other ie:

    A_n = Sum (from i=1 to n) of:
    1/sqrt(i)

    B_n = sqrt(n)

    That A_(n+1) / A_n > B_(n+1) / B_n

    I have tried squaring both sides and simplifying the expression. I just cannot seem to make any progress on this. Any help would be appreciated.
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 19, 2009 #2

    Office_Shredder

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    Product or sum? If it's the product of 1/sqrt(i) then each time n increases, the product decreases (and is always less than 1)
     
  4. Sep 19, 2009 #3
    I'm sorry it should be a sum, I'll edit the original post.
     
  5. Sep 20, 2009 #4
    Anyone have a hint to even start he problem? I'm a bit stumped.
     
  6. Sep 21, 2009 #5
    Can you think of the sum as a Riemann sum that is greater than an easy integral?
     
  7. Sep 21, 2009 #6
    If anyone cares I ended up doing this by induction.

    n=2 trivial to check

    A_n = Sum (from i=1 to n): 1/sqrt(i)

    A_(n+1) = A_n + 1/sqrt(n+1)

    Must show: A_(n+1) > sqrt(n+1) = sqrt(n) + sqrt(n+1) - sqrt(n)

    by our induction hypothesis this is if and only if:

    1/sqrt(n+1) > sqrt(n+1) - sqrt(n)

    iff

    1 > n + 1 - sqrt(n^2 + n)

    Note: sqrt(n^2 + n) > n thus our above inequality holds proving

    A_(n+1) > sqrt(n+1)
     
  8. Sep 29, 2009 #7
    If you are familiar with the AM-HM inequality, then you could do this:

    [tex]\frac{\sum^{n}_{i=1} \sqrt{i}}{n} \ge \frac{n}{\sum^{n}_{i=1} \frac{1}{\sqrt{i}}}[/tex]

    So

    [tex]\sum^{n}_{i = 1} \frac{1}{\sqrt{i}} \ge \frac{n^2}{\sum^{n}_{i=1} \sqrt{i}} \ge \frac{n^2}{\sum^{n}_{i=1} \sqrt{n}} = \frac{n^2}{n \sqrt{n}} = \sqrt{n} [/tex]

    Equality occurs iff each of the square-roots in the original sum is equal, which is only the case for n = 1. In all other cases, the inequality is strict.
     
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