19
$\begingroup$

$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$

With a first look this must give $1$ as a result but have a problem to explain it.

How can I do it?

Edit

I noticed that it is $\frac{\infty}{\infty}$.

$\lim_{n \to \infty}{n^{n}\frac{(\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1)}{n^n}}= \lim_{n \to \infty}{\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1}=1$

Is this correct?

  • 0
    So you say that the actual result is $0\cdot\infty+1$ and this is undefined.2011-10-29

4 Answers 4

22

We can write $(1^1+2^2+\cdots+n^n)/n^n$ as $a_n + b_n + 1$, where $ a_n = \frac{1^1+2^2+\cdots+(n-2)^{n-2}}{n^n} \text{ and } b_n = \frac{(n-1)^{n-1}}{n^n}. $ Both $a_n$ and $b_n$ are positive, and also $ a_n < \frac{(n-2)(n-2)^{n-2}}{n^n} < b_n < \frac{n^{n-1}}{n^n} = \frac1n. $ The squeeze theorem should allow you to prove that your answer of 1 is correct.

  • 1
    @Chris: since we want the quantity $a_n+b_n+1$ to tend to $1$ in the limit, it's important to have a diminishing upper bound like b_n < 1/n.2012-05-28
18

Stolz-Cesàro is probably overkill, but solves the problem easily.

The limit

$\lim\limits_{n\to\infty} \frac{(n+1)^{n+1}}{(n+1)^{n+1}-n^n} =1 \,,$ is very simple to calculate.

14

Let $f(n) = (1^1 + 2^2 + 3^3 + \cdots + n^n)/n^n$. You want to show $\lim_{n \to \infty} f(n) = 1$.

It's obvious that $f(n) > 1$ for all $n$.

For an upper bound, $ f(n) \le {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} + {n^{n-1} \over n^n} + {n^n \over n^n} = {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} + {1 \over n} + 1.$

Now I leave it to you to find some bound $g(n)$, with $\lim_{n \to \infty} g(n) = 0$ and $ {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} \le g(n). $ So you have $ 1 < f(n) < 1 + {1 \over n} + g(n) $ and apply the squeeze theorem to finish the proof.

  • 1
    ha, you beat me by 45 seconds!2011-10-29
3

I know I'm terribly late, but I'd like to post a similar but a bit different from Micheal and Greg's approach. Plese tell me in the comments if it's not sufficiently different.

Let $a_n=\frac{1^1+2^2+3^3+\cdots+(n-1)^{n-1}}{n^n}. $ So we have $a_{n+1}=\frac{1^1+2^2+3^3+\cdots+(n-1)^{n-1}+n^n}{(n+1)^{n+1}}=\frac{n^n(a_n+1)}{(n+1)^{n+1}}$ and thus $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{a_nn^n}{(n+1)^{n+1}}+\frac{n^n}{(n+1)^{n+1}}=\lim_{n\to\infty}\frac{a_nn^n}{(n+1)^{n+1}}.\tag{1}$ Since $\lim_{n\to\infty}\frac{n^n}{(n+1)^{n+1}}=0,$ $(1)$ yields that $a_n$ either converges to $0$ or diverges to $+\infty$. But $\lim_{n\to\infty}a_n\le\lim_{n\to\infty}\frac{\overbrace{(n-1)^{n-1}+(n-1)^{n-1}+\cdots+(n-1)^{n-1}}^{n-1 \ \text{times}}}{n^n}=\lim_{n\to\infty}\frac{(n-1)^n}{n^n}=e^{-1}, $ hence $\lim_{n\to\infty}a_n=0,$ which allows us to conclude $\lim_{n\to\infty}\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}=1.$