Determining whether the sequence converges or diverges as n--> infinity

  • Context: MHB 
  • Thread starter Thread starter shamieh
  • Start date Start date
  • Tags Tags
    Infinity Sequence
Click For Summary

Discussion Overview

The discussion revolves around determining the convergence or divergence of specific sequences as \( n \to \infty \). Participants analyze three sequences, providing their reasoning and justifications for their conclusions, which include both mathematical techniques and intuitive reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that the sequence \( a_n = 2 + (-1)^n \) diverges due to oscillation.
  • Others propose that the sequence \( a_n = \frac{n}{e^n} \) converges to 0, using L'Hôpital's rule to justify their reasoning.
  • There is uncertainty regarding the sequence \( a_n = (1 + \frac{2}{n})^n \), with some participants suggesting it diverges while others believe it converges to \( e^2 \). One participant expresses confusion about whether it converges to 1 or diverges.

Areas of Agreement / Disagreement

Participants generally agree on the divergence of the first sequence and the convergence of the second to 0. However, there is disagreement and uncertainty regarding the third sequence, with competing views on whether it converges to \( e^2 \) or diverges.

Contextual Notes

Participants reference L'Hôpital's rule and the definition of \( e \) in their arguments, indicating that the discussion may depend on specific mathematical definitions and assumptions that are not fully explored.

shamieh
Messages
538
Reaction score
0
Can someone check my solutions?

Do the following sequences $${a_n}$$ converges or diverge as$$ \n\to\infty$$? If a sequence converges find its limit. Justify your answers.

1. $$a_n = 2 +(-1)^n$$

Answer: so can I say that as lim n --> infinity the sequence diverges by oscillation?

2. $$a_n = \frac{n}{e^n}$$

Answer: so: using l'opitals$$\frac{n}{e^n} = \frac{1}{e^n} = \frac{1}{\infty} = 0$$

$$\therefore$$ sequence converges to 0 ?

3. $$a_n = (1 + \frac{2}{n})^n$$

Answer: So using lopital I got $$(1)^n$$ so would that mean that the sequence diverges becase its 1 + 1 + 1 + 1 to infinity...
 
Physics news on Phys.org
shamieh said:
Can someone check my solutions?

Do the following sequences $${a_n}$$ converges or diverge as$$ \n\to\infty$$? If a sequence converges find its limit. Justify your answers.

1. $$a_n = 2 +(-1)^n$$

Answer: so can I say that as lim n --> infinity the sequence diverges by oscillation?

2. $$a_n = \frac{n}{e^n}$$

Answer: so: using l'opitals$$\frac{n}{e^n} = \frac{1}{e^n} = \frac{1}{\infty} = 0$$

$$\therefore$$ sequence converges to 0 ?

3. $$a_n = (1 + \frac{2}{n})^n$$

Answer: So using lopital I got $$(1)^n$$ so would that mean that the sequence diverges becase its 1 + 1 + 1 + 1 to infinity...

These are fine provided you do not have to prove from first principles
 
so is my #3 correct? or will it converge to 1 since its a sequence?
 
shamieh said:
so is my #3 correct? or will it converge to 1 since its a sequence?

I fear I've been writing rubbish. The series converges, in fact it is $e^2$. How you prove that depends on your definition of $e$
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K