Question

Determine whether the sequence is converges or diverges. if converges find the limit.

1) (e^n+e^-n )/ (n^(2n) - 1)

2)an=cos(2/n)

Answer 1

As we are dealing with sequences, we need only to find the limit. If the limit is a real number, then it converges. The second one should be pretty easy, right? :-) 2/n goes to zero, so cos(2/n) goes to?

Answer 2

is converger but the first is confusing

Answer 3

Indeed. I agree with you. The limit seems to go to zero, but to prove it looks hard.

Answer 4

that second one is 1 so we need to find the limit
but idk about the first one

Answer 5

Hmm, to show that it's decrescent seems easy, but can we prove that it's bounded? But still, we would have to show the limit calculation. Let me think.

Answer 6

sure

Answer 7

Well, something that comes to mind is that: \[\lim_{n \rightarrow \infty} \frac{e^n + e^{-n}}{n^{2n} - 1} = \lim_{n \rightarrow \infty} ( \frac{e^n}{n^{2n} - 1} + \frac{e^{-n}}{n^{2n} -1})\]The second limit is trivial

Answer 8

We need to show that (e^n)/(n^(2n) - 1) has a limit.

Answer 9

ok

Answer 10

you uses the partial fraction

Answer 11

I don't see a clear way of proving the first limit using rigor. I would say that n^(2n) > n^n, and n^n > e^n for n >= 3. So, n^(2n) > e^n at least for n >= 3. As n grows larger, the bottom grows faster than the top, so the limit goes to 0.

Answer 12

so the first problem is converges and the second too ?

Answer 13

Yup, both do. :-)

Answer 14

The first limit is 0, the second is 1 (cos(0)), so both are real numbers and the sequences converge.

Answer 15

then we need to find the limit of both question