Yes, I meant c>0. Thank you for clarifying!

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    Divergence
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Homework Help Overview

The discussion revolves around the convergence of the sequence 2^(n^2)/n! and its behavior as n approaches infinity. Participants are exploring the implications of limits and the relationship between sequences in the context of divergence and infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • One participant suggests comparing 2^(n^2)/n! to 2^(n^2)/n^n to demonstrate divergence, hinting at the use of logarithms for analysis. Another participant questions the conditions under which the product of two sequences approaches infinity, specifically discussing the implications of the limit of one sequence being infinite while the other remains bounded.

Discussion Status

The discussion is active, with participants providing hints and exploring different interpretations of limits and divergence. There is no explicit consensus, but some guidance has been offered regarding the comparison of sequences and the conditions for limits.

Contextual Notes

Participants are considering the implications of the limit of a bounded sequence in relation to an unbounded sequence, with specific attention to the case where the bounded sequence approaches a positive constant versus zero.

peripatein
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Hi,
How may I show that 2^(n^2)/n! converges to infinity?
 
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peripatein said:
Hi,
How may I show that 2^(n^2)/n! converges to infinity?

That is SO divergent you can afford to be pretty sloppy. 2^(n^2)/n!>2^(n^2)/n^n, right? So show 2^(n^2)/n^n diverges. Hint: look at the log.
 
Thank you, Dick!
 
Would it be correct to say that if for sequences a_n and b_n, lim a_n = infinity and |b_n|< c < infinity, then lim|a_n*b_n| = infinity?
(I think it should be correct, as we may infer that lim |bn| = c and then the limit of the product of a_n and b_n would yield c*infinity which is always infinity.)
 
peripatein said:
Would it be correct to say that if for sequences a_n and b_n, lim a_n = infinity and |b_n|< c < infinity, then lim|a_n*b_n| = infinity?
(I think it should be correct, as we may infer that lim |bn| = c and then the limit of the product of a_n and b_n would yield c*infinity which is always infinity.)

If you mean lim |b_n|=c with c>0, then sure. If c=0, then you need to think more.
 
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