Convergence of an: e^(n)sin(n) & e^(2n)/[4^n]

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Discussion Overview

The discussion focuses on the convergence or divergence of two sequences defined as an = e^(n) sin(n) and an = e^(2n)/[4^n]. Participants explore methods for determining convergence, including numerical evaluation and theoretical approaches.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant requests assistance in determining the convergence of the sequences.
  • Another participant questions what determines convergence and asks about tests for convergence.
  • A participant suggests evaluating the sequences by plugging in values for n to identify patterns as n approaches infinity.
  • For the first sequence, one participant expresses a belief that it diverges based on their evaluation.
  • For the second sequence, a participant proposes rewriting it as an = (E/4)^n, where E = e^2, and argues that since E/4 > 1, the sequence diverges.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of the sequences, with some suggesting divergence while others propose methods for analysis without reaching a consensus.

Contextual Notes

Participants have not fully resolved the mathematical steps or assumptions underlying their evaluations, and the discussion remains exploratory regarding the convergence tests applicable to the sequences.

hytuoc
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Would someone please help me w/ the problems below? Thanks so much
Determine whether the sequence converges or diverges, n, if it converges, find lim as n approaches infinity of "an" (subscript "n")
1) an= e^(n) sin(n)
2) an = e^(2n)/ [4^n]
 
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What determines whether a series converges, or not ?

(what tests are there to check for convergence ?)
 
more like finding a derivative of those
 
hytuoc said:
Would someone please help me w/ the problems below? Thanks so much
Determine whether the sequence converges or diverges, n, if it converges, find lim as n approaches infinity of "an" (subscript "n")
1) an= e^(n) sin(n)
2) an = e^(2n)/ [4^n]
I'm not sure if you titled this thread right; this is a sequence question but that's ok.

One way to get a handle on convergence is to plug in a few values for n and see if there is a pattern in the higher n (as n approaches infinity). Try this for 1) and see what happens. I think it diverges.

For 2) you can do the same thing and/or rewrite it first into
E=e^2
an=(E/4)^n
then you should have a theorem which states that since E/4>1, the sequence diverges.
 

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