Proof by induction

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  • #1
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Homework Statement



Prove this by induction

n
∑(1/√n) > 2(√(n+1) -1)
i=1


The Attempt at a Solution



I have tried for several hours, with no real progress. When i assume that it works for n=k, and then try to prove for n=k+1 i get 2(√(k+2)-1)... i don't know how to partition this expression... can someone point me in the right direction....
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Prove this by induction

n
∑(1/√n) > 2(√(n+1) -1)
i=1


The Attempt at a Solution



I have tried for several hours, with no real progress. When i assume that it works for n=k, and then try to prove for n=k+1 i get 2(√(k+2)-1)... i don't know how to partition this expression... can someone point me in the right direction....

Are you absolutely required to use induction? If not, there is a quick proof using elementary calculus (specifically, definite integration).
 
  • #3
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The task is part of the induction chapter in the textbook.
 
  • #4
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Are you sure that the left-hand side of the inequality is...

[itex] \sum_{k=1}^{n} \frac 1{\sqrt {n}}[/itex], rather than [itex] \sum_{k=1}^{n} \frac 1{\sqrt {k}}[/itex] ?
 
  • #5
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1
Are you sure that the left-hand side of the inequality is...

[itex] \sum_{k=1}^{n} \frac 1{\sqrt {n}}[/itex], rather than [itex] \sum_{k=1}^{n} \frac 1{\sqrt {k}}[/itex] ?

woops!, i can correct it...
 
  • #6
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m
∑(1/√n) > 2(√(n+1) -1)
n=1
 
  • #7
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1
Again, the problem for me lies in that i can't partition(divide in to a sum) the expression on the right...
 
  • #9
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So sorry, I'm not so good at the different placements for the different sum-limits/-startpoints. I first introduced "k" when i assumed that the inequality actually was valid, and then i set n=k+1 and tried to prove it...

This should be right;
n
∑(1/√i) > 2(√(n+1) -1)
i=1
 
  • #11
120
6
I think you meant to write...

[itex] \sum_{n=1}^{m} \frac 1{\sqrt{n}} > \frac 2{\sqrt{m+1}} -1 [/itex].

If this is the case, here's a way to start the inductive step:

[itex]\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = ( \sum_{n=1}^{m} \frac 1{\sqrt{n}}) + \frac 1{\sqrt{m+1}} [/itex]
 
  • #12
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1
no, it is 2(√(n+1) -1) = 2√(n+1) -2
 
  • #13
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Let me know if you'd like more hints, but I'm pretty sure you can finish the problem from here ;)
 
Last edited by a moderator:
  • #14
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6
no, it is 2(√(n+1) -1) = 2√(n+1) -2


Oh... the right hand side is a product. Oops.

That's ok. The steps in the proof won't change much.
 
  • #15
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1
[itex]\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = ( \sum_{n=1}^{m} \frac 1{\sqrt{n}}) + \frac 1{\sqrt{m+1}} [/itex]
I already knew this, but how am i supposed to compare this result to the right side? the expression on the right side is impossible to write as a sum of any sort...? At least : 2√(m+2)...
 
  • #16
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6
Starting from the Left-hand side of the inequality you wish to prove (P(m+1)),

[itex] \sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = \sum_{n=1}^{m} \frac 1{\sqrt{n}} + \frac 1{\sqrt{m+1}} [/itex].

Recall that you assumed P(m) to be true, that is

[itex] \sum_{n=1}^{m} \frac 1{\sqrt{n}} > 2 \cdot ((\sqrt{m+1}) -1) [/itex],

So recognizing this, you have...

[itex] \sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = \sum_{n=1}^{m} \frac 1{\sqrt{n}} + \frac 1{\sqrt{m+1}} > 2 \cdot ((\sqrt{m+1}) -1) + \frac 1{\sqrt{m+1}} [/itex].
 
  • #17
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1
If i contract the right side i get: (2m +3 -2√(m+1))/√(m+1)

Don't see how this is gonna help.. :)
 
  • #18
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1
dont i have to show that (2m +3 -2√(m+1))/√(m+1) can be written as 2(√(m+2) -1) to prove it...?

or maybe not since its an inequality...?
 
Last edited:
  • #19
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6
dont i have to show that (2m +3 -2√(m+1))/√(m+1) can be written as 2(√(m+2) -1) to prove it...?

or maybe not since its an inequality...?

If you plug in some numbers you'll see...

[itex]2 \cdot (\sqrt{m+1} - 1) + \frac 1{\sqrt{m+1}} \neq 2 \cdot (\sqrt {(m+1)+1} - 1) [/itex].

You're next step should be to show that...

[itex]2 \cdot (\sqrt{m+1} - 1) + \frac 1{\sqrt{m+1}} > 2 \cdot (\sqrt {(m+1)+1} - 1) [/itex].
 
  • #20
Ray Vickson
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[itex]\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = ( \sum_{n=1}^{m} \frac 1{\sqrt{n}}) + \frac 1{\sqrt{m+1}} [/itex]
I already knew this, but how am i supposed to compare this result to the right side? the expression on the right side is impossible to write as a sum of any sort...? At least : 2√(m+2)...

Your problem is to show that
[tex] 2 \sqrt{n+1}-2 + \frac{1}{\sqrt{n+1}} \geq 2 \sqrt{n+2} - 2,\\
\text{or}\\
\sqrt{n+2} - \sqrt{n+1} \leq \frac{1}{2 \sqrt{n+1}}[/tex]
If I were doing it I would write
[tex] \sqrt{n+2} = \sqrt{n+1} \left( 1 + \frac{1}{n+1} \right)^{1/2}[/tex]
and go on from there.
 
  • #21
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yes, i got 9/4 > 8/4! or 9>8 thanks!, now its just about proving the basecase and its finished?
 
  • #22
Ray Vickson
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yes, i got 9/4 > 8/4! or 9>8 thanks!, now its just about proving the basecase and its finished?

Which message are you responding to? Please use the "Quote" button, so that the discussion can be followed accurately.
 
  • #23
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Which message are you responding to? Please use the "Quote" button, so that the discussion can be followed accurately.

both of them :) the inequality that both of you posted implies that 9>8. Since this is true, the inequality you posted must be true ?
 
  • #24
Ray Vickson
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both of them :) the inequality that both of you posted implies that 9>8. Since this is true, the inequality you posted must be true ?

I have no idea what you are talking about; we need to know something for all n (or at least, for all n larger than, say, 2 or 3), so 8 and 9 have nothing to do with the problem.
 
  • #25
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both of them :) the inequality that both of you posted implies that 9>8. Since this is true, the inequality you posted must be true ?

We've invested a lot of time helping you out, so we would like to see you get the right answer. Could you post how you came to this conclusion?
 

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