Comparison test for convergence

Thread starter tony873004
Start date Dec 9, 2007
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Comparison Comparison test Convergence Test

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SUMMARY

The forum discussion centers on the convergence of the series \(\sum\limits_{n = 2}^\infty {\frac{{\sqrt n }}{{n - 1}}}\). The user demonstrates that the series diverges by comparing it to the p-series \(\frac{1}{{n^{1/2}}}\) with \(p = \frac{1}{2}\), which is known to diverge. The Limit Comparison Test is also applied to reinforce the conclusion of divergence, highlighting the necessity of this additional step for rigor in mathematical proofs.

PREREQUISITES

Understanding of series convergence tests, specifically the Limit Comparison Test.
Familiarity with p-series and their convergence criteria.
Basic knowledge of limits and asymptotic behavior of functions.
Ability to manipulate algebraic expressions involving limits.

NEXT STEPS

Study the Limit Comparison Test in detail, including its conditions and applications.
Explore the properties of p-series and their convergence for different values of \(p\).
Learn about other convergence tests such as the Ratio Test and the Root Test.
Practice solving various series convergence problems to solidify understanding.

USEFUL FOR

Students studying calculus, particularly those focusing on series and convergence, as well as educators looking to clarify the application of convergence tests in mathematical proofs.

Dec 9, 2007

tony873004

Science Advisor

Gold Member

Homework Statement

\sum\limits_{n = 2}^\infty {\frac{{\sqrt n }}{{n - 1}}}

Homework Equations

The Attempt at a Solution

\begin{array}{l} a_n = \frac{{\sqrt n }}{{n - 1}},\,\,b_n = \frac{{\sqrt n }}{n} = \frac{{n^{1/2} }}{n} = \frac{1}{{n^{1/2} }}\,\,{\rm{is_ a_ p - series_ with }}p{\rm{ = }}\frac{1}{2},\,\frac{1}{2} < 1\,{\rm{diverges}} \\ \\ \frac{{\sqrt n }}{{n - 1}} > \frac{{\sqrt n }}{n}\, \\ \\ \end{array}
so it is divergent.

But in this example from the notes, the teacher used the Limit Comparison Test, taking the \mathop {\lim }\limits_{n \to \infty } \frac{{a_n }}{{b_n }}, and using this to justify that since b_nis is divergent, therefore a_nis also divergent. I understand this logic, but why was the extra step necessary when the comparison test alone showed it to be divergent?