# Finding the limit of a complicated funcion

1. Jan 24, 2012

### xcrunner448

1. The problem statement, all variables and given/known data

Find $\stackrel{lim}{_{n\rightarrow\infty}}\frac{1^{1}+2^{2}+...+(n-1)^{n-1}+n^{n}}{n^{n}}$.

2. The attempt at a solution

At first I split up the fraction into a sum of a bunch of terms, and said that all of the terms went to 0 except the last, which is 1. But then I realized that in the limit, there is an infinite number of terms so it makes no sense to say that the "last term" is 1. I think I found a way to prove it by finding a recursive formula for the function inside the limit and then using that to get upper and lower bounds, which both converge to 1. But that was really long and seemed like a very round-about way of doing things. Is there a simpler way to get the answer?

2. Jan 24, 2012

### Dick

Sure. You do a really ruthless upper bound of the first (n-2) terms of your series. But I find it hard to believe you found a recursive formula for the function that gives you anything useful. Can you show me that?

3. Jan 25, 2012

### xcrunner448

Ok, here's my current solution. Like I said, it seems like an overly complicated way of doing it, and I wouldn't be surprised if there is some mistake in there somewhere.

#### Attached Files:

• ###### solution.pdf
File size:
33 KB
Views:
76
4. Jan 25, 2012

### Dick

Maybe that works. It does seem a little complicated. I'd start by saying $1^{1}+2 ^{2}+...+(n-2)^{n-2} \le (n-2) (n-2)^{n-2}$. Do you see how it would go from there?

5. Jan 25, 2012

### xcrunner448

Ok, that makes sense. It makes it a sum of 3 terms, the first two of which go to 0 and the last goes to 1. I think I tried that too, but with the first n-1 terms and that didn't work. I guess it never occurred to me to try it with the first n-2 terms. Thanks for your help.

6. Jan 25, 2012

### Dick

Yeah. Doing it with the first (n-1) terms gives an upper limit that's too big. So you need to chose a smaller upper limit.