Divergent Series Proof

  • #1

Homework Statement


Prove that the series diverges: [itex]\sum_{i=1}^{\infty}\sqrt{n+1}-\sqrt{n}[/itex]


The Attempt at a Solution


I'm trying to use the comparison test, but I don't know what to compare it to. All I have done so far is change the terms to be in fraction form:
[itex]\sqrt{n+1}[/itex]-[itex]\sqrt{n}[/itex]=[itex]\frac{1}{\sqrt{n+1}+\sqrt{n}}[/itex]

Any clues on what to do next would be great. Thanks!
 

Answers and Replies

  • #2
Dick
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Homework Statement


Prove that the series diverges: [itex]\sum_{i=1}^{\infty}\sqrt{n+1}-\sqrt{n}[/itex]


The Attempt at a Solution


I'm trying to use the comparison test, but I don't know what to compare it to. All I have done so far is change the terms to be in fraction form:
[itex]\sqrt{n+1}[/itex]-[itex]\sqrt{n}[/itex]=[itex]\frac{1}{\sqrt{n+1}+\sqrt{n}}[/itex]

Any clues on what to do next would be great. Thanks!
Try comparing with [itex]\frac{1}{\sqrt{n+1}+\sqrt{n+1}}[/itex]. Does that converge or diverge? How is it related to your original series?
 
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  • #3
Ray Vickson
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Homework Statement


Prove that the series diverges: [itex]\sum_{i=1}^{\infty}\sqrt{n+1}-\sqrt{n}[/itex]


The Attempt at a Solution


I'm trying to use the comparison test, but I don't know what to compare it to. All I have done so far is change the terms to be in fraction form:
[itex]\sqrt{n+1}[/itex]-[itex]\sqrt{n}[/itex]=[itex]\frac{1}{\sqrt{n+1}+\sqrt{n}}[/itex]

Any clues on what to do next would be great. Thanks!
Look at the behavior of ## t_n \equiv \sqrt{n+1} - \sqrt{n}## for large ##n##. It helps to write
[tex] \sqrt{n+1} = \sqrt{n} \left( 1 + \frac{1}{n} \right)^{1/2} [/tex]
 

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