# What is the limit $\lim_{n\to\infty}\frac{1^n+2^n+3^n+\dots+n^n}{n^{n+1}}$?

I want to determine the limit given below: $$\lim_{n\to\infty}\frac{1^n+2^n+3^n+\ldots+n^n}{n^{n+1}}$$ I have tried to solve thise several times ,but with no results.I have tried using lema stolz cezaro and managed to find a general formula for the summation, but couldnt prove it $$\frac{1}{n}\sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^n$$ Edit: Thank you for your responses,but i want to ask you if it is possible to solve the problem by not using integrals :D. NOTEIm in the eleventh grade and i had been given this problem by my math tutor ,so, as a consequence i dont know how to interpretate the great majority of your answers.Thanks for your time ,gents!

You only need to exploit the identity (proved here): $$\sum_{k=0}^{K}\binom{n+k}{n}=\binom{n+K+1}{n+1}$$ and the trivial bound: $$m!\binom{n}{m} \leq n^m \leq m!\binom{n+m}{m}$$ to have: $$\forall m\in\mathbb{N},\quad \lim_{n\to +\infty}\frac{1^m+2^m+\ldots+n^m}{n^{m+1}}=\frac{1}{m+1}$$ from which it follows that your limit is zero.

As an alternative approach, since over $[0,1]$ we have $1-x\leq e^{-x}$:

$$\frac{1}{n}\sum_{k=0}^{n}\left(1-\frac{k}{n}\right)^n \leq \frac{1}{n}\sum_{k=0}^{n}e^{-k}\leq\frac{1}{n}\sum_{k=0}^{+\infty}e^{-k}=\frac{e}{n(e-1)}.$$

• I wanted to ask you soemthing.For the last relation presented when k=n-1 it is 0^infinity.I have solved it the same way and i was told it has some problems. – Razvan Paraschiv Apr 21 '15 at 15:23
• @RazvanParaschiv: my last line does not involve any limit, it is just an inequality holding for every $n\geq 1$. The trick is to consider limits just after that and state that the limit is zero by squeezing, since $$\lim_{n\to +\infty}\frac{e}{n(e-1)}=0,$$ no issues. – Jack D'Aurizio Apr 21 '15 at 15:28
• Thank you! I apologize for my miscomprehension – Razvan Paraschiv Apr 21 '15 at 15:29
• Jack: does 'from which it follows that your limit is zero' actually follow? After all, $\lim_{n\to\infty}(1+\frac1n)^m=1$ for all $m$, but that doesn't imply the same result when $m$ is replaced by the limiting variable. Isn't this the exact same situation? – Steven Stadnicki Apr 21 '15 at 15:34
• @StevenStadnicki: well, in that case one may prove that $$f(m,n) = \frac{1^m+\ldots+n^m}{n^{m+1}}$$ is monotonic with respect to both $m$ and $n$ and exploit this fact to be allowed to exchange limits. – Jack D'Aurizio Apr 21 '15 at 15:37

Note that $\sum_{j=1}^n j^n \le \int_{1}^{n+1} x^n$. (And recall that $((n+1)/n)^{n}$ has a finite limit.)

• Thanks. Yours has the advanateg of remaining more discrete, which I generally like. – quid Apr 21 '15 at 15:06
• I like it too. I think I have managed to find another proof, similar to yours, and without integrals at all. Just see below. – Jack D'Aurizio Apr 21 '15 at 15:11
• @JackD'Aurizio I see in my browser above :D – RE60K Apr 25 '15 at 15:37

Here's another point of view. For each $n\in\mathbb N$ choose an integer $k(n)\in\{0,1,\ldots,n\}$ such that the following two properties are satisfied: $$\lim_{n\to\infty}\frac{k(n)}{n}=1\qquad\text{and}\qquad\lim_{n\to\infty}\left(\frac{k(n)}{n}\right)^n=0.\tag{\ast}$$ Some possible choices are: $$k(n)=\lfloor n-\log{n}\rfloor\qquad\text{or}\qquad k(n)=\lfloor n-\sqrt n\rfloor\qquad\text{or}\qquad k(n)=\lfloor n^{\frac{n}{n+1}}\rfloor,$$ where $\lfloor x\rfloor$ is the largest integer not exceeding $x$, i.e. the floor function. (That $(\ast)$ holds follows e.g. from $\lim_{n\to\infty}(1+\frac1n)^n=e$ in the first two cases and from $\lim_{n\to\infty}n^{\frac1n}=1$ in the third case.)

Given such a sequence $k(n)$, we have the following bounds: \begin{align}0&\leq\frac{1^n+2^n+\ldots+k(n)^n+(k(n)+1)^n+\ldots+n^n}{n^{n+1}}\\&\leq\frac{k(n)k(n)^n+(n-k(n))n^n}{n^{n+1}}\\&=\left(\frac{k(n)}{n}\right)^{n+1}+1-\frac{k(n)}{n}.\end{align} This last sequence converges to zero by $(*)$, so our sequence is bounded between two sequences converging to zero. We may conclude that $$\lim_{n\to\infty}\frac{1^n+2^n+3^n+\ldots+n^n}{n^{n+1}}=0.$$

• that is the "razor" function? – Razvan Paraschiv Apr 25 '15 at 14:19
• @RazvanParaschiv: Yes, I guess you could call it that. I should probably remark that the main part of this proof (requiring most of the work) consists of finding such a function and proving its properties, i.e. $(\ast)$. If you have any further questions about this, ask and I'll clarify. – Dejan Govc Apr 25 '15 at 14:43
• the square brakets signify the full part of their content? – Razvan Paraschiv Apr 25 '15 at 14:47
• @RazvanParaschiv: you mean the whole part? Yes, $\lfloor x\rfloor$ is defined as the unique integer $k$ such that $k\leq x<k+1$. See here and here. – Dejan Govc Apr 25 '15 at 15:00
• thank you sir indeed ! in my country the notation si the following x<=[x]<=x+1 – Razvan Paraschiv Apr 25 '15 at 15:57