What's the Best Way to Evaluate This Limit?

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Discussion Overview

The discussion revolves around evaluating the limit of the integral \(\lim_{n\rightarrow\infty}\int_0^1 e^{t^n}dt\). Participants explore various approaches and techniques to tackle this limit, including integral tricks and series expansions.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how to start evaluating the limit and seeks guidance on potential approaches.
  • Another participant suggests that \(e^{t^n}\) approaches \(e^0=1\) on the interval \([0,d)\) for any \(0 < d < 1\) and proposes taking the limit as \(d\) approaches 1.
  • A different approach is presented using a series expansion of \(e^{t^n}\) as \(\sum_{i=0}^{\infty}\frac{t^{ni}}{i!}\), leading to a manipulation of the integral and summation, which is claimed to converge to 1 as \(n\) increases.
  • One participant questions whether the limit converges to 0 instead of 1 and reflects on the possibility of changing the limits of integration, indicating a consideration of the Taylor series for further insights.
  • A later reply corrects the previous statement, asserting that the limit indeed converges to 1.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the limit's value, with some suggesting it converges to 1 while others propose it converges to 0. The discussion remains unresolved regarding the final outcome.

Contextual Notes

There are limitations in the assumptions made about the behavior of the integral and the convergence of the series, which are not fully explored or resolved in the discussion.

Philcorp
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[tex]\lim_{n\rightarrow\infty}\int_0^1 e^{t^n}dt[/tex]. I am not really sure where to start to evaluate this limit, but I probably have enough tricks up my sleeve to solve it if someone knowledgeable is able to point me in the right direction. My usual integral tricks seem to fail here.

Cheers.
 
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I think you can get [tex]e^{t^n}[/tex] as close as you want to [tex]e^0=1[/tex] on [0,d), for any 0 < d < 1. Then take the limit as d -> 1.
 
You might do it this way:
[tex]e^{t^{n}}=\sum_{i=0}^{\infty}\frac{t^{ni}}{i!}[/tex]
Thus, switching about the summation limit and the integral (acceptable for any fixed n), we get:
[tex]\int_{0}^{1}e^{t^{n}}dt=\sum_{i=0}^{\infty}\frac{t^{ni+1}}{(ni+1)i!}\mid_{0}^{1}=\sum_{i=0}^{1}\frac{1}{(ni+1)i!}[/tex]
This should converge to 1 as n jumps into the air,
since for i>=1, we have:
[tex]\frac{1}{(ni+1)i!}<\frac{1}{n}*\frac{1}{i!}[/tex]
 
Last edited:
Cool, thanks for the help! Also, I think you mean it converges to 0? The replies have made me wonder about changing the limits on the integral to be from a to b, since it seems that the 0 to 1 case converges quite nicely, I should have thought to do a taylor series :blushing:, oops, and I am sure there is some nice upper limit for which this converges.

Thanks again!Oops, nope it is 1!
 
Last edited:

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