Proving Divergence of the Series: \sum_{i=1}^{\infty}\sqrt{n+1}-\sqrt{n}

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SUMMARY

The series \(\sum_{i=1}^{\infty}(\sqrt{n+1}-\sqrt{n})\) diverges. The transformation \(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\) is crucial for applying the comparison test. By comparing \(\frac{1}{\sqrt{n+1}+\sqrt{n}}\) with a known divergent series, one can establish the divergence of the original series. Analyzing the behavior of \(t_n \equiv \sqrt{n+1} - \sqrt{n}\) as \(n\) approaches infinity confirms this conclusion.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the comparison test in calculus
  • Knowledge of limits and asymptotic behavior
  • Basic algebraic manipulation of square roots
NEXT STEPS
  • Study the comparison test for series convergence in detail
  • Learn about the behavior of sequences and series at infinity
  • Explore the concept of asymptotic analysis in calculus
  • Investigate other divergence tests, such as the ratio test and root test
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Students studying calculus, particularly those focusing on series and sequences, as well as educators looking for examples of divergence proofs.

analysis001
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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!
 
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analysis001 said:

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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analysis001 said:

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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