Determine whether series is convergent or divergent

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SUMMARY

The series \(\sum^{\infty}_{n=1} \frac{1}{\sqrt{n+1}+\sqrt{n}}\) is determined to be divergent by comparing it to the p-series \(\sum^{\infty}_{n=1} \frac{1}{\sqrt{n}}\), which is known to diverge. The rationalization approach suggested in the discussion is not necessary; instead, evaluating the limit of the function as \(n\) approaches infinity reveals that it behaves like \(\frac{1}{2\sqrt{n}}\). This confirms the divergence of the original series.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with p-series and their properties
  • Knowledge of limits and asymptotic behavior
  • Basic algebraic manipulation techniques, including rationalization
NEXT STEPS
  • Study the properties of p-series and their convergence criteria
  • Learn about the Limit Comparison Test for series
  • Explore techniques for evaluating limits of series as \(n\) approaches infinity
  • Review rationalization methods and their appropriate applications in series analysis
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Students studying calculus, particularly those focusing on series convergence, mathematicians, and educators looking to enhance their understanding of series analysis techniques.

Nan1teZ
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Homework Statement



Determine whether or not the series \sum^{\infty}_{n=1} \frac{1}{\sqrt{n+1}+\sqrt{n}} converges.

The Attempt at a Solution



Assuming this diverges, I rationalize it to get get \sum^{\infty}_{n=1} \sqrt{n+1} - \sqrt{n}. How would I proceed further?

Is this even the right approach?
 
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Well, you can't just assume it diverges. I haven't studied series in quite a while. The rationalizing of the denominators seems like a promising approach, but someone else will have to comment on that.

I think the following is a valid method. Let f(n) be the expression we are summing. As n-> +inf f(n) -> 1/(2sqrt(n)). But the infinite series with 1/sqrt(n) is a p-series and I think it's divergent.
 
snipez90 is right. Don't 'rationalize' it. Just compare it with a divergent p-series.
 

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