# Simple way to get a complex Limit?

## Main Question or Discussion Point

I'm helping a few people out with school maths and I came across this limit (in connection with the Ratio Test):

$$\lim_{n \to \infty}\left(\frac{n}{n+1}\right)^n$$

now I can figure out the answer to this myself using l'Hôpital's Rule but the student's I'm teaching wouldn't have come across it yet, so it's either a misprint in their book - or there's a simple way of figuring it out which I can't see. Anyone have any ideas?

could we look at the limit of $$(\left(\frac{n}{n+1}\right)^n - \left(\frac{(n-1)}{(n-1)+1}\right)^{n-1} )$$ ?

I mean if $$\lim_{n \to \infty} (a_{n} - a_{n-1}) \to \infty$$ then doesn't that tell us something?

Last edited:
Sorry, I probably shouldn't have mentioned the ratio test at all, I'm just looking to find the limit, which equals $$e^{-1}$$ which I got by taking the natural log of the term above and applying l'Hôpital's Rule.

you want to use the squeeze theorem

OK, any ideas of what two functions to pick?

Mute
Homework Helper
Surely they have come across the definition of $e$? If so, it's simple enough:

$$\left(\frac{n}{n+1}\right)^n = \left(\frac{n+1}{n}\right)^{-n} = \left[\left(1 + \frac{1}{n}\right)^n\right]^{-1}$$

Taking the limit of the last expression and using the definition of $e$ gives $e^{-1}$.

Is there a reason why they wouldn't be allowed to do it this way?

mute is correct

Excellent, yes they would have come across that definition of $e$. But as for myself on the other hand...let's just say it's been a while.

Thanks Mute and grmnsplx.