# Limit of a sequence with n as a variable, and to the power of n

1. Nov 17, 2008

### CrystalEyes

1. The problem statement, all variables and given/known data
Find the limits of the following sequences, if they exist. Justify your answers.

{($$\frac{n-1}{n}$$ )$$^{n}$$}$$^{\infty}_{n=1}$$ or maybe $$\sum^{\infty}_{n=1}(\frac{n-1}{n} )^{n}$$ is clearer?

2. Relevant equations
I know the whole part that it's a sequence doesn't really matter, that i'm just finding the limit as n $$\rightarrow$$$$\infty$$ but that's not so easy...

3. The attempt at a solution
I know that the limit of that (sorry i'm too lazy to retype it) equals the [limit (as n approaches infinity) of ((n-1)/n) ] ^n by the "limit laws" if you want to call them that (the whole think to the nth degree including the limit, sorry if this is really confusing, i'm new to the equation insert thing) but i'm kind of stuck. Because if you evaluate just ((n-1)/n) you end up with infinity over infinity, and it's not something to some other variable that the power is being taken, it's to the same variable, "n", which should be going to infinity anyways, so I'm quite stuck! please help? and thanks!!

2. Nov 17, 2008

### lurflurf

[(n-1)/n]^n=1/[1+1/(n-1)]^n
a well known limit since for large n
1+1/(n-1)~exp[1/(n-1)]

3. Nov 17, 2008

### Dick

The limit has the form 1^infinity. Can't you use l'Hopital's rule on it? If you find the limit exists and is nonzero you don't have to worry about the series, it will diverge.

4. Nov 17, 2008

### CrystalEyes

I didn't think i could because the whole thing is to the power of n, but i guess i wasn't thinking straight? Anyways, thanks again for the help I was just studying for my calc2 exam tomorrow and came across this one

5. Nov 17, 2008

### Dick

If you take the log, you can turn it into a 0/0 form that you can apply l'Hopital to. Or if you already know that lim (1+x/n)^n=e^x, you can rearrange it into that form.