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

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

How can I do it?


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?

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I improved the TeXing a bit. It seemed that you were missing $n \to \infty$ in the question, so I took the liberty to add it. – Srivatsan Oct 29 '11 at 19:57
Why is the last limit 1? You have a 1 and $n-1$ terms which converge to 0, so when you take the limit you get one + infinitely many zeros. The answer can still be anything.... – N. S. Oct 29 '11 at 21:32
So you say that the actual result is $0\cdot\infty+1$ and this is undefined. – ertzan topal Oct 29 '11 at 21:38
up vote 21 down vote accepted

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.

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We can't just apply del Hospital rule? It is $\frac{\infty}{\infty}$. – ertzan topal Oct 29 '11 at 20:23
What's the derivative of the numerator as $n \to \infty$? To find that you'd need to find a way to interpret the numerator when $n$ is not an integer. – Michael Lugo Oct 29 '11 at 20:47
@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 $b_n < 1/n$. – Greg Martin May 28 '12 at 22:26

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.

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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.

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ha, you beat me by 45 seconds! – Greg Martin Oct 29 '11 at 20:02

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

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