Question

Help with e^-1

Answer 1

I have this function
\[\left| \frac{ n^{(n+1)^2} }{ (n+1)^{(n+1)^2} }*\frac{ n^{n^2} }{ (1-n)^{n^2} } \right|\]
and I know its goes to e^-1 when n --> ∞
But have can I show it?

Answer 2

I'm playing with it now, maybe you can get it to look like the definition of e in the denominator with a little clever algebraic manipulation.

e=(1+1/n)^n

just for reference.

Answer 3

I know that definition.. :)

Answer 4

Cool, now make it work for you! Try factoring that out or something haha, I don't know myself.

Answer 5

WOW I'm tired giving bad advice, sorry. Let's just separate out the limit into pieces and multiply them together when we're done. First:

\[\lim_{n \rightarrow \infty}\frac{ n }{ n+1 } ^{(n+1)^2} \]

Ok try to take this limit and you see that it becomes 1^infinity which is an indeterminate form, so we're free to use L'Hopital's rule.

\[ e^{\lim_{n \rightarrow \infty}\frac{\ln(\frac{ n }{ n+1 })}{(n+1)^{-2}}} \]

Now when you differentiate them you'll get a piece of your limit. I need to go to bed, not sure if that was very helpful either.

Answer 6

@Kainui thant you.. Sleep well:)