Find $ \lim _{n\to \infty} \frac{n^1+\dots+n^n}{1^n+\dots+n^n}$

Question

What is the limit of:

$$\lim_{n\to \infty} \frac{n^1+\dots+n^n}{1^n+\dots+n^n} = ?$$

I did some computation with big numbers, I guess it is in the interval $\left(\frac{1}{2},1\right)$.

Answer 1

The answer is $1-1/e$. In fact, one may prove $$\lim_{n\to\infty}\frac{n^1+n^2+\ldots+n^n}{n^n}=1$$ and $$\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}=\frac{1}{1-1/e}$$ and the conclusion follows.

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The numerator is a geometric series that evaluates to $(n^{n+1}-n)/(n-1)$. One thus have \begin{align} \lim_{n\to\infty}\frac{n^1+n^2+\ldots+n^n}{n^n}=\lim_{n\to\infty}\frac{1-n^{-n-1}}{1-1/n}=1 \end{align} The denominator can be estimated in a quick way, by monotone convergence theorem or, in case this theorem is not available, by the following argument $$\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}>\lim_{n\to\infty}\sum_{k=0}^K\left(\frac{n-k}n\right)^n=\sum_{k=0}^Ke^{-k}$$ Let $k\to\infty$, we have $$\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}\ge\frac{1}{1-1/e}$$ On the other hand, since $\ln(1+x)\le x$, we have \begin{align} \lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}&=\lim_{n\to\infty}\sum_{k=0}^{n-1}\left(1-\frac kn\right)^n\\ &=\lim_{n\to\infty}\sum_{k=0}^{n-1}\exp(n\ln(1-k/n))\\ &\le\lim_{n\to\infty}\sum_{k=0}^{n-1}\exp(-k)\\ &=\frac1{1-1/e} \end{align}