Limits and sequences math problem

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Homework Help Overview

The problem involves finding the limit of a sequence defined by a_n = (2n)! * 2^(2n) / (n!)^2 * (2n + 1) * 5^(2n) as n approaches infinity. The context is within the subject area of limits and sequences in mathematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the ratio test and its application to sequences versus series. Some express uncertainty about the appropriateness of the ratio test for this specific problem, while others attempt to evaluate the limit directly.

Discussion Status

The discussion is ongoing, with participants sharing different perspectives on the use of the ratio test and the evaluation of the limit. There is no explicit consensus, but various approaches are being explored, including direct evaluation and the ratio test.

Contextual Notes

Some participants note a lack of familiarity with using the ratio test on sequences, suggesting it may not be commonly taught or referenced in their learning materials.

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Homework Statement


Find the limit as n[tex]\rightarrow[/tex][tex]\infty[/tex] of the sequence

an=[tex]\frac{(2n)! 2<sup>2n</sup>}{(n!)<sup>2</sup> (2n+1) 5<sup>2n</sup>}[/tex]

Homework Equations





The Attempt at a Solution

 
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Fixed the formatting, I think:

[tex] a_n = \frac{(2n)! 2^{2n} }{(n!)^2 (2n+1) 5^{2n}}.[/tex]

If that's right, what have you tried?
 


I've just worked out how to do the problem. I found the ratio of an+1 to an, as n approaches infinity, and the ratio is less than 1 (16/25), so by the ratio test, this means the sequence approaches 0 as n approaches infinity. Thanks for fixing the formatting. Indeed the expression is correct. :smile:
 


The ratio test is to be used on terms of an infinite series, [itex]\sum a_n[/itex]. If the limit of the ratio a_(n + 1)/a_n is less than 1, the series converges. You should be able to evaluate the limit of this sequence directly.

The tricky part is evaluating (2n)!/[(n!)^2 (2n + 1)], since 2^(2n)/5^(2n) = (2/5)^(2n) --> 0 as n --> infinity.
 


Mark44 said:
The ratio test is to be used on terms of an infinite series, [itex]\sum a_n[/itex]. If the limit of the ratio a_(n + 1)/a_n is less than 1, the series converges. You should be able to evaluate the limit of this sequence directly.

The tricky part is evaluating (2n)!/[(n!)^2 (2n + 1)], since 2^(2n)/5^(2n) = (2/5)^(2n) --> 0 as n --> infinity.

You can also use the ratio test on a sequence. If |a_(n+1)/a_n| goes to a limit less than 1 then the sequence converges to zero. As it must if the series is to converge. For some reason this doesn't seem to be taught much as a method.
 


Dick said:
You can also use the ratio test on a sequence. If |a_(n+1)/a_n| goes to a limit less than 1 then the sequence converges to zero. As it must if the series is to converge. For some reason this doesn't seem to be taught much as a method.

I don't remember learning about using the ratio test on a sequence, or seeing it in any of the calculus texts I taught out of.
 


Mark44 said:
I don't remember learning about using the ratio test on a sequence, or seeing it in any of the calculus texts I taught out of.

No, I don't think I have either. But it works, right?
 

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