Evaluate $$\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$$

My solution

Notice that $$\lim_{n \to \infty}\frac{n+n^2+n^3+\cdots +n^n}{n^n}=\lim_{n \to \infty}\frac{n(n^n-1)}{(n-1)n^n}=\lim_{n \to \infty}\frac{1-\dfrac{1}{n^n}}{1-\dfrac{1}{n}}=1,$$and

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

Hence,\begin{align*}\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}&=\lim_{n \to \infty}\frac{\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=\frac{\lim\limits_{n \to \infty}\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\lim\limits_{n \to \infty}\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=1-\frac{1}{e}.\end{align*}

The solution posted above need to quote an uncommon limit. Is there another more simple and more direct solution?

  • $\begingroup$ Wolfram alpha answer to the series is 1 wolframalpha.com/input/… $\endgroup$ – emil May 23 '18 at 6:22
  • 5
    $\begingroup$ I'm afraid this's not true, though I don't know the reason for that. $\endgroup$ – mengdie1982 May 23 '18 at 6:52
  • $\begingroup$ The computer is not as smart as we are? $\endgroup$ – mengdie1982 May 23 '18 at 11:45
  • 4
    $\begingroup$ No, the computer is pretty dumb, though pretty fast. $\endgroup$ – Ivan Neretin May 23 '18 at 11:57
  • $\begingroup$ With Stoltz-Cesaro theorem its easy $\endgroup$ – Milan Jul 20 at 13:22

We shall prove that $$ \frac{1^n+2^n+\cdots+n^n}{n^n}\to \frac{\mathrm{e}}{\mathrm{e}-1}\tag{$\star$} $$ First of all, $\log (1-x)<-x$, for all $x\in(0,1)$ and hence $$ \log\left(1-\frac{k}{n}\right)<-\frac{k}{n}\quad\Longrightarrow\quad \left(1-\frac{k}{n}\right)^n<\mathrm{e}^{-k}, \quad \text{for all $n>k$} $$ and thus $$ \frac{1^n+2^n+\cdots+n^n}{n^n}=\sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^n <\sum_{k=0}^{n-1}\mathrm{e}^{-k}<\sum_{k=0}^{\infty}\mathrm{e}^{-k}=\frac{1}{1-\frac{1}{\mathrm{e}}}=\frac{\mathrm{e}}{\mathrm{e}-1}. $$ Hence $$ \limsup_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^n}\le \frac{\mathrm{e}}{\mathrm{e}-1}. \tag{1} $$

Meanwhile, for all $k\in\mathbb N$, $$ \frac{(n-k)^n}{n^n}=\left(1-\frac{k}{n}\right)^n\to\mathrm{e}^{-k}, $$ and hence, for every $k\in\mathbb N$ fixed, $$ \frac{1^n+2^n+\cdots+n^n}{n^n}\ge \frac{(n-k)^n+(n-k+1)^n+\cdots+n^n}{n^n}\\=\left(1-\frac{k}{n}\right)^n+\left(1-\frac{k-1}{n}\right)^n+\cdots+\left(1-\frac{1}{n}\right)^n+1\to \mathrm{e}^{-k} +\mathrm{e}^{-k+1}+\cdots+1=\frac{\mathrm{e}-\mathrm{e}^{-k}}{\mathrm{e-1}}. $$ Hence, for all $k\in\mathbb N$, $$ \liminf_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^n}\ge \frac{\mathrm{e}-\mathrm{e}^{-k}}{\mathrm{e-1}} $$ and thus $$ \liminf_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^n}\ge \sup_{k\in\mathbb N}\frac{\mathrm{e}-\mathrm{e}^{-k}}{\mathrm{e-1}}=\frac{\mathrm{e}}{\mathrm{e-1}} \tag{2} $$ Combining $(1)$ & $(2)$, we obtain $(\star)$.


A Query to @Teddy38's Answer

Yes, we can agree that $$n+n^2+\cdots+n^n=\frac{n}{n-1}(n^n-1)$$ and $$\frac{n^{n+1}}{n+1}=\int_{0}^{n}x^n\ dx<1^n+2^n+\cdots+n^n<\int_{1}^{n+1}x^n\ dx=\frac{(n+1)^{n+1}-1}{n+1}.$$ Thus,we obtain $$\frac{n+1}{n-1}\cdot \frac{n^{n+1}-n}{(n+1)^{n+1}-1}<\dfrac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}<\frac{n+1}{n-1}\cdot\frac{n^n-1}{n^n}.$$Now, let's evaluate the limits of the both sides as $n \to \infty.$ For the left side, we have $$\lim_{n \to \infty} \dfrac{n+1}{n-1} \cdot \lim_{n \to \infty}\dfrac{1-\dfrac{1}{n^n}}{\left(1+\dfrac{1}{n}\right)^n\cdot\left(1+\dfrac{1}{n}\right)-\dfrac{1}{n^{n+1}}}=\frac{1}{e}.$$ As for the right side, we have $$\lim_{n \to \infty} \dfrac{n+1}{n-1} \cdot \lim_{n \to \infty}\left(1-\dfrac{1}{n^n}\right)=1.$$

From the two aspects, we may only conclude that, $$\varliminf_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}\geq\frac{1}{e},$$and $$\varlimsup_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}\leq 1.$$ That's to say, if the limit we want really exists, then $$\frac{1}{e}\leq\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n} \leq 1.$$This is true, but can't give the accurate value of the limit we want.

  • $\begingroup$ Yup, my comment was too optimistic. I'll delete it. $\endgroup$ – Teddy38 May 24 '18 at 7:35
  • $\begingroup$ @Teddy38 It doesn't matter. Thanks for your participation in the discussion. $\endgroup$ – mengdie1982 May 24 '18 at 8:16

I think the above sandwiching methods are only feasible when you already know the result before tackling the problem. I will present a method starting from scratch.

Note that for a special case of Faulhaber’s formula, the numerator equals $$S=\sum^n_{k=1}k^n=n^n\sum^n_{k=0}\frac{B^+_k}{k!}\frac{n!n^{1-k}}{(n-k+1)!}$$

It is well known that $$\frac{n!n^a}{(n+a)!}=1+o(\cdots)$$ as $n\to\infty$.

Thus, the required limit $$\frac{S}{n^n}\~ \sum^n_{k=0}\frac{B^+_k}{k!}=\frac1{1-e^{-1}}=\color{RED}{\frac{e}{e-1}}$$ by noting the generating function of Bernoulli numbers $$\frac{t}{1-e^{-t}}=\sum^\infty_{k=0}\frac{B^+_kt^k}{k!}$$

p.s. I don’t know what the $o(\cdots)$ should be. Anyone who knows it please edit my answer.

Also, I don’t know how to type the symbol for asymptotic in Mathjax, please help me...

  • $\begingroup$ Symbol for asymptotics is, surprisingly, \asymp $\asymp$ $\endgroup$ – Yuriy S May 24 '18 at 11:07
  • $\begingroup$ @YuriyS But can’t we type ‘~’? $\endgroup$ – Szeto May 26 '18 at 0:12
  • $\begingroup$ @Szeto, try \approx. $\endgroup$ – Jose Arnaldo Bebita-Dris May 30 '18 at 5:14

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