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Induction Problem

  1. May 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove by induction. The sum from 1 to n of [tex]\frac{1}{\sqrt{i}}[/tex] [tex]\geq[/tex] [tex]\sqrt{n}[/tex].

    2. Relevant equations

    3. The attempt at a solution
    I verified the base case, and all of that, I just can't get to anything useful. I tried expanding the sum and adding the [tex]\frac{1}{\sqrt{n+1}}[/tex] but it didn't come to anything useful.

  2. jcsd
  3. May 9, 2008 #2
    What have you gotten after adding the [itex]1/\sqrt{n+1}[/itex] ? What would you like the numerator of the fraction to be after combining?
  4. May 10, 2008 #3
    after the induction assumption you have to prove the inequality:
    [tex]\sum_{i=1}^n {\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}[/tex]
    given that [tex]\sum_{i=1}^n {\frac{1}{\sqrt{i}}} \geq \sqrt{n}[/tex]
    Last edited: May 10, 2008
  5. May 10, 2008 #4


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    Assuming [tex]\sum_{i=1}^n {\frac{1}{\sqrt{i}}} \geq \sqrt{n}[/tex]. [tex]\sum_{i=1}^n {\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n}+ \frac{1}{\sqrt{n+1}}[/tex]
    Adding those, [tex]\sqrt{n}+ \frac{1}{\sqrt{n+1}}=\frac{\sqrt{n^2+ n}+1}{\sqrt{n+1}}[/tex]
    that's what you want if [tex]\sqrt{n^2+ n}+ 1\ge \sqrt{n+1}[/tex] and that should be easy to prove.
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