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

In summary, the conversation is about determining whether given sequences converge or diverge and finding the limit as n approaches infinity of the sequences. The first sequence, an=e^(n)sin(n), is suggested to diverge while the second sequence, an=e^(2n)/[4^n], is shown to diverge using a theorem.
  • #1
hytuoc
26
0
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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  • #2
What determines whether a series converges, or not ?

(what tests are there to check for convergence ?)
 
  • #3
more like finding a derivative of those
 
  • #4
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.
 

1. What is the convergence of e^(n)sin(n)?

The sequence e^(n)sin(n) diverges, meaning it does not approach a finite limit as n approaches infinity. This can be seen by considering the oscillating behavior of the sine function, which cancels out the exponential growth of e^(n).

2. Does e^(2n)/[4^n] converge or diverge?

The sequence e^(2n)/[4^n] converges to 0 as n approaches infinity. This can be seen by writing the expression as (e^2)^n/(4^n), which can be simplified to (e^2/4)^n. Since e^2/4 is less than 1, the sequence approaches 0 as n becomes larger and larger.

3. Is the convergence of e^(n)sin(n) affected by changing the value of n?

Yes, the convergence of e^(n)sin(n) is affected by changing the value of n. As n increases, the sine function will cause the sequence to oscillate more, making it diverge. However, if n is a very large negative number, the sequence will approach 0 since the exponential growth of e^(n) will dominate over the oscillating behavior of the sine function.

4. How can the convergence of e^(n)sin(n) be proven?

The convergence of e^(n)sin(n) can be proven using the limit comparison test. By comparing it to a known convergent sequence, such as e^n or 1/n, it can be shown that e^(n)sin(n) also converges. Additionally, the alternating series test can also be used to prove convergence by showing that the sequence is alternating and the terms decrease in magnitude.

5. Can the convergence of e^(2n)/[4^n] be determined without using a convergence test?

Yes, the convergence of e^(2n)/[4^n] can be determined by using basic properties of exponents. By writing the expression as (e^2)^n/(4^n), it is clear that the base of the exponential term is less than 1, meaning the sequence approaches 0 as n becomes larger. Therefore, no convergence test is necessary to determine the convergence of this sequence.

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