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

In summary, the conversation discusses three sequences and whether they converge or diverge as n approaches infinity. The first sequence, a_n = 2 + (-1)^n, is determined to diverge by oscillation. The second sequence, a_n = n/e^n, is found to converge to 0 using L'Hopital's rule. The third sequence, a_n = (1 + 2/n)^n, is initially thought to diverge but is later corrected to converge to e^2. The method of proving the convergence depends on the definition of e.
  • #1
shamieh
539
0
Can someone check my solutions?

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

1. \(\displaystyle a_n = 2 +(-1)^n\)

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

2. \(\displaystyle a_n = \frac{n}{e^n}\)

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

\(\displaystyle \therefore\) sequence converges to 0 ?

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

Answer: So using lopital I got \(\displaystyle (1)^n\) so would that mean that the sequence diverges becase its 1 + 1 + 1 + 1 to infinity...
 
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  • #2
shamieh said:
Can someone check my solutions?

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

1. \(\displaystyle a_n = 2 +(-1)^n\)

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

2. \(\displaystyle a_n = \frac{n}{e^n}\)

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

\(\displaystyle \therefore\) sequence converges to 0 ?

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

Answer: So using lopital I got \(\displaystyle (1)^n\) 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
 
  • #3
so is my #3 correct? or will it converge to 1 since its a sequence?
 
  • #4
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:

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

What does it mean for a sequence to converge?

A sequence converges if its terms approach a single fixed value as the index, n, increases. This fixed value is known as the limit of the sequence.

How do you determine if a sequence converges?

To determine if a sequence converges, you can evaluate its limit by taking the limit of the sequence as n approaches infinity. If the limit exists and is a finite number, then the sequence converges.

What is the difference between a convergent and a divergent sequence?

A convergent sequence has a limit that exists and is a finite number, while a divergent sequence does not have a limit or has a limit that is infinite.

What if the limit of a sequence does not exist?

If the limit of a sequence does not exist, then the sequence is considered to be divergent. This could happen if the terms in the sequence do not approach a single fixed value as n increases.

Can a sequence converge to more than one limit?

No, a sequence can only converge to one limit. If the sequence has multiple limits, then it is considered to be divergent.

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