Convergence of \sum_{n=1}^\infty \frac{1}{n!}: A Basic Comparison Test

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Homework Help Overview

The discussion revolves around the convergence of the series \(\sum_{n=1}^\infty \frac{1}{n!}\) and the comparison of another series \(\sum_{n=1}^\infty \frac{2^n}{n^2}\). Participants are exploring methods to establish convergence or divergence through comparison tests.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about what to compare the series \(\sum_{n=1}^\infty \frac{1}{n!}\) to, suggesting a p-series as a potential comparison. Other participants suggest comparing it to \(1/n^2\). Additionally, the original poster raises a new series \(\sum_{n=1}^\infty \frac{2^n}{n^2}\) and questions what it should be compared to, noting an assumption about its divergence.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for comparisons and discussing the implications of those comparisons. There is a recognition of the need for further clarification on how to apply comparison tests effectively. Some participants express frustration with the constraints of the exercise.

Contextual Notes

There is mention of the limit test and its applicability, with some participants questioning the restrictions placed on the methods allowed for proving convergence or divergence.

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[tex]\sum_{n=1}^\infty \frac{1}{n!}[/tex]

I understand what n! means, but I have no clue what to compare this to. It is obvious to me that the sum converges, but I'm not sure how to prove it. I assume I would compare it to a p-series but I need help. Thanks!
 
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Compare it to [itex]1/n^2[/itex]...
 
micromass said:
Compare it to [itex]1/n^2[/itex]...

lol, I just realized how simple this is. my bad.
 
Sorry, one more.

[tex]\sum_{n=1}^\infty \frac{2^n}{n^2}[/tex]

What would I compare this to?

I can clearly see that it diverges since numerator is waaaay bigger but I don't know how to prove it.
 
Calculate the limit of the terms and show that the limit isn't 0.
 
micromass said:
Calculate the limit of the terms and show that the limit isn't 0.

Sorry I didn't specify. I understand how to use the limit test. For this problem I am supposed to compare it to something. Thanks for putting up with my questions :).

edit: since 2^n is soo much bigger then n^2 can I compare it to 2^n?
 
Maybe the harmonic series??

It's a stupid exercise anyway if you're not allowed to do the limit test.
 

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