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Induction Problem

  • Thread starter dtl42
  • Start date
  • #1
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Homework Statement


Prove by induction. The sum from 1 to n of [tex]\frac{1}{\sqrt{i}}[/tex] [tex]\geq[/tex] [tex]\sqrt{n}[/tex].

Homework Equations


none.

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.

Thanks
 

Answers and Replies

  • #2
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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?
 
  • #3
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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:
  • #4
HallsofIvy
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Homework Helper
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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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