Question

Sum of:
(-1)^(n+1)*n^(-2)*n^(-n+4)*(1+3/n)^(n^2)
with n=1 to +inifinity.
Need to tell if the series converges or diverges, don't need to know the specific number it goes to. I've done the root test on it but got stuck when it came to dealing with the exponents. Please help, thanks

Answer 1

\((-1)^{n+1}\) -- Okay, it alternates, so we need to establish only that the terms converge to zero. Then we're done. Well, assuming it converges.

\(\dfrac{\left(1+\dfrac{3}{n}\right)^{n^2}}{n^{2}\cdot n^{n-4}}\)

Hmmm... What do you think? \(\lim\limits_{x\rightarrow\infty}\left(1+\dfrac{3}{n}\right)^{n} = e^{3}\). Does that get us anywhere?

Answer 2

So that limit you show at the bottom of your response is a rule right? So e^3 would be the numerator, then you can combine the denominator

Answer 3

After you rewrite the equation, is it then that you do the root test?

Answer 4

No, e^3 is not the numerator. There's an n^2 up there, not just an n.

Answer 5

I did the root test and it brought the n^2 down to an n

Answer 6

Maybe compare to \(\dfrac{e^{3n}}{n^{n-2}}\).

Do we get l'Hospital?

Answer 7

We definitely get 1^infinity for the numerator, and 1 for the denominator

Answer 8

That's no good. They are \(\dfrac{\infty}{\infty}\). Can we use it?

Answer 9

i thought l'hopitals rule could also be 1^infinity?

Answer 10

At least my professor says it can...

Answer 11

It can, but that's not what we have.

Answer 12

okay, so infinity/infinity, i figured out why i thought the other way

Answer 13

if you have \[\frac{ \infty }{\infty}\]
then you would take the derivative of each the top and bottom

Answer 14

Yeah, no. That won't get us anywhere. We need to rewrite to understand.

\(\left(\dfrac{\left(1+\dfrac{3}{n}\right)^{n}}{n}\right)^{n}\cdot n^{2} < \left(\dfrac{e}{n}\right)^{n}\cdot n^{2} = \dfrac{e^{n}}{n^{n-2}}\) and that clearly converges to zero after n > 2.