How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$?

Question

I can show that the following limit exists but I am having difficulties to find it. It is $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}$$ Can someone please help me?

Answer 1

An asymptotic expansion can be obtained as below. More terms can be included by using more terms in the expansions of $\exp$ and $\log$. $$ \begin{align} \sum_{k=0}^n\frac{k^n}{n^n} &=\sum_{k=0}^n\left(1-\frac{k}{n}\right)^n\\ &=\sum_{k=0}^n\exp\left(n\log\left(1-\frac{k}{n}\right)\right)\\ &=\sum_{k=0}^{\sqrt{n}}\exp\left(n\log\left(1-\frac{k}{n}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\ &=\sum_{k=0}^{\sqrt{n}}\exp\left(-k-\frac{1}{2n}k^2+O\left(\frac{k^3}{n^2}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\ &=\sum_{k=0}^{\sqrt{n}}e^{-k}\exp\left(-\frac{1}{2n}k^2+O\left(\frac{k^3}{n^2}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\ &=\sum_{k=0}^{\sqrt{n}}e^{-k}\left(1-\frac{1}{2n}k^2+O\left(\frac{k^ 4}{n^2}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\ &=\sum_{k=0}^{\sqrt{n}}e^{-k}-\frac{1}{2n}\sum_{k=0}^{\sqrt{n}}k^2e^{-k}+O\left(\frac{1}{n^2}\right)\\ &=\frac{e}{e-1}-\frac{1}{2n}\frac{e(e+1)}{(e-1)^3}+O\left(\frac{1}{n^2}\right) \end{align} $$ Several steps use $$ \sum_{k=n}^\infty e^{-k}k^m=O(e^{-n}n^m) $$ which decays faster than any power of $n$.

Answer 2

Just for reference: With aid of some fancy theorem, you can skip most of hard analysis. As in other answers, we begin by writing

$$ \sum_{k=1}^{n} \left( \frac{k}{n}\right)^n \ \overset{k \to n-k}{=} \ \sum_{k=0}^{n-1} \left( 1 - \frac{k}{n}\right)^n \ = \ \sum_{k=0}^{\infty} \left( 1 - \frac{k}{n}\right)^n \mathbf{1}_{\{k < n\}}, $$

where $\mathbf{1}_{\{k < n\}}$ is the indicator function which takes value $1$ if $k < n$ and $0$ otherwise. Now for each $0 \leq k < n$, utilizing the inequality $\log(1-x) \leq -x$ which holds for all $x \in [0,1)$ shows that

$$ \left( 1 - \frac{k}{n}\right)^n = e^{n \log(1 - \frac{k}{n})} \leq e^{-k}. $$

Since $\sum_{k=0}^{\infty} e^{-k} < \infty$, by the dominated convergence theorem we can interchange the infinite sum and the limit:

$$ \lim_{n\to\infty} \sum_{k=1}^{n} \left( \frac{k}{n}\right)^n = \sum_{k=0}^{\infty} \lim_{n\to\infty} \left( 1 - \frac{k}{n}\right)^n \mathbf{1}_{\{k < n\}} = \sum_{k=0}^{\infty} e^{-k} = \frac{1}{1 - e^{-1}}. $$

Answer 3

Finally, I have suffered this proof. Consider functions $$ f_n(x)=\left(1-\frac{\lfloor x\rfloor}{n}\right)^n\chi_{[0,n+1]}(x) $$ Note that $$ \int\limits_{[0,+\infty)} f_n(x)d\mu(x)=\sum\limits_{k=0}^n\int\limits_{[k,k+1)}\left(1-\frac{\lfloor x\rfloor}{n}\right)^nd\mu(x)= \sum\limits_{k=0}^n\left(1-\frac{k}{n}\right)^n $$ $$ \lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}\left(1-\frac{\lfloor x\rfloor}{n}\right)^n\cdot \lim\limits_{n\to\infty}\chi_{[0,n+1]}(x)=e^{\lfloor x\rfloor} $$ One may check that $\{f_n:n\in\mathbb{N}\}$ is a non-decreasing sequence of non-negative functions, then using monotone convergence theorem we get $$ \lim\limits_{n\to\infty}\sum\limits_{k=0}^n\left(\frac{k}{n}\right)^n= \lim\limits_{n\to\infty}\sum\limits_{k=0}^n\left(1-\frac{k}{n}\right)^n= \lim\limits_{n\to\infty}\int\limits_{[0,+\infty)} f_n(x)d\mu(x)= $$ $$ \int\limits_{[0,+\infty)} \lim\limits_{n\to\infty}f_n(x)d\mu(x)= \int\limits_{[0,+\infty)} e^{\lfloor x\rfloor}d\mu(x)= \sum\limits_{k=0}^\infty e^{-k}=\frac{1}{1-e^{-1}} $$

Answer 4

Let's notice a few things. All the terms are positive, bounded between $0$ and $1$, and there is a term that is exactly $1$. What about the next largest term?

So we ask ourselves what $\lim \limits_{n \to \infty} \left( \dfrac{n-1}{n} \right)^n$ is, and after a little calculation we see that this limit is $1/e$. The 'next' term involves $\lim \limits_{n \to \infty} \left( \dfrac{n-2}{n} \right)^n = e^{-2}$. So heuristically, we would expect the limit to be

$$1 + e^{-1} + e^{-2} + \dots = \frac{1}{1-\frac{1}{e}}$$

Working only a little harder, you can justify that this is the limit.

Answer 5

$\sum_{k=1}^n(k/n)^n=\sum_{0<k\le n}(1-k/n)^n$, and let $a_k(n)=(1-k/n)^n$. For $0<k\le n^{1/3}$, we have $$\ln a_k(n)=n\ln\left(1-\frac kn\right)=-n\left(\frac kn+O\left(\frac kn\right)\right)=-k+O\left(\frac{k^2}n\right)$$ thus $$a_k(n)=e^{-k}\left(1+O\left(\frac{k^2}n\right)\right)$$ Let $b_k(n)=e^{-k}$, $c_k(n)=k^2e^{-k}/n$, we have $a_k(n)=b_k(n)+O(c_k(n))$ over $0<k\le n^{1/3}$. Thus, we have $$\sum_{0<k\le n}a_k(n)=\sum_{k>0}b_k(n)+O(\Sigma_a(n))+O(\Sigma_b(n))+O(\Sigma_c(n))$$ where $$\sum_{k>0}b_k(n)=\sum_{k>0}e^{-k}=\frac e{e-1}$$ and \begin{align*} \Sigma_b(n)&=\sum_{k>n^{1/3}}e^{-k}=O(e^{n^{1/3}})\\ \Sigma_a(n)&=\sum_{n^{1/3}<k\le n}\left(1-\frac kn\right)^n\le\sum_{n^{1/3}<k\le n}e^{-k}=O(e^{n^{1/3}})\\ \Sigma_c(n)&=\sum_{0<k\le n^{1/3}}e^{-k}k^2/n\le\sum_{k>0}e^{-k}k^2/n=O\left(\frac 1n\right) \end{align*} Hence, we have $\sum_{0<k\le n}(1-k/n)^n=e/(e-1)+O(1/n)$.

Can anybody give a more accurate approximation? The key to the approximation is to find the asymptotics for $\sum_{k>0}\exp(-k-k^2/2n)$, like the Bell sum $\sum_{k>0}e^{-k^2/n}$.

Edit anon pointed out that it's theta function: $\sum_ke^{-(k+t)^2/n}$, so the Fourier series works pretty well for the asymptotics: $$\Theta_n(t)=\sqrt{\pi n}\left(1+2e^{-\pi^2 n}(\cos2\pi t)+2e^{-4\pi^2 n}(\cos4\pi t)+2e^{-9\pi^2 n}(\cos6\pi t)+\cdots\right)$$ But I have no idea about Fourier series because I know very little about calculus!

Answer 6

First, we can rewrite the formula:

$$\frac{\sum_{i=1}^{n}i^n}{n^n} = \sum_{i=1}^{n}\frac{i^n}{n^n} = \sum_{i=0}^{n-1}\frac{(n-i)^n}{n^n} = \sum_{i=0}^{n-1}\left(1-\frac{i}{n} \right)^n$$

We can easily prove the following inequality relations:

$$\frac{1}{\text{e}^i} - \frac{i^2}{n\text{e}^i} \le \frac{1}{\text{e}^i}\left(1-\frac{i}{n} \right)^i\le \left(1-\frac{i}{n} \right)^n \le \frac{1}{\text{e}^i} $$

Therefore, we can get

$$\sum_{i=0}^{n-1}\frac{1}{\text{e}^i}-\frac{1}{n}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i}\le\sum_{i=0}^{n-1}\left(1-\frac{i}{n} \right)^n \le \sum_{i=0}^{n-1}\frac{1}{\text{e}^i}$$

So

$$\lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{1}{\text{e}^i}-\lim_{n\to \infty}\frac{1}{n}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i}\le\lim_{n\to \infty}\sum_{i=0}^{n-1}\left(1-\frac{i}{n} \right)^n \le \lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{1}{\text{e}^i}$$

We can easily know that

$$\lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{1}{\text{e}^i} = \frac{\text{e}}{\text{e}-1}$$

$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i} = \left(\lim_{n\to \infty}\frac{1}{n}\right)\cdot \left(\lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i}\right) = 0 \cdot \frac{\text{e}(\text{e}+1)}{(\text{e}-1)^3} = 0$$

Finally, we can get the answer: $$\lim_{n\to \infty} \frac{\sum_{i=1}^{n}i^n}{n^n} = \lim_{n\to \infty}\sum_{i=0}^{n-1}\left(1-\frac{i}{n} \right)^n = \frac{\text{e}}{\text{e}-1}$$

For more details, you can watch my video on Bilibili, if you have learned Chinese.

Here is my video link: https://www.bilibili.com/video/BV1yG411g7K1/