Yesterday, a friend asked me the following problem. Prove that: \lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n}=\frac{e}{e-1} I will briefly record the solution process here.
Solution
First, it can be noted that when n is sufficiently large, \frac{1^n + 2^n +\dots + n^n}{n^n}=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\dots+\left(\frac{n}{n}\right)^n the main terms are concentrated in the last few terms. Therefore, we can calculate it in reverse order: \begin{aligned}\frac{1^n + 2^n +\dots + n^n}{n^n}=&\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\dots+\left(\frac{n}{n}\right)^n\\ =&\left(\frac{n}{n}\right)^n+\dots+\left(\frac{2}{n}\right)^n+\left(\frac{1}{n}\right)^n\end{aligned} Since we have \left(\frac{n-i}{n}\right)^n=\left(1-\frac{i}{n}\right)^n which is an increasing function with respect to n, it follows that \left(1-\frac{i}{n}\right)^n \leq \lim_{n\to\infty} \left(1-\frac{i}{n}\right)^n = e^{-i} Thus, the inequality \left(\frac{n}{n}\right)^n+\dots+\left(\frac{2}{n}\right)^n+\left(\frac{1}{n}\right)^n \leq e^0 + e^{-1}+e^{-2}+\dots=\frac{e}{e-1} holds universally.
On the other hand, the main task is to find a lower bound estimate for \left(\frac{n-i}{n}\right)^n. We have \begin{aligned}&\ln\left[\left(\frac{n-i}{n}\right)^n\right]\\ =&n\ln\left(1-\frac{i}{n}\right)\\ =&n\left[-\frac{i}{n}-\frac{1}{2}\left(\frac{i}{n}\right)^2-\frac{1}{3}\left(\frac{i}{n}\right)^3-\dots\right]\end{aligned}
Note that we have the inequality \ln(1-x) = -x -\frac{1}{2}x^2-\frac{1}{3}x^3-\dots > -x-x^2. Of course, this inequality does not hold universally, but it holds when x is small. We can roughly estimate that it holds for 0\leq x \leq \frac{1}{2}. Therefore, when i \leq \frac{n}{2}, we have \ln\left[\left(\frac{n-i}{n}\right)^n\right]\geq n\left[-\frac{i}{n}-\left(\frac{i}{n}\right)^2\right]=-i-\frac{i^2}{n} or \left(\frac{n-i}{n}\right)^n \geq e^{-i-i^2/n}\geq e^{-i} \left(1-\frac{i^2}{n}\right)\geq e^{-i} -\frac{i^2}{n}
Now, we know that \sum_{i=0}^m i^2 \sim m^3. Thus, we only take the part where i < n^{1/4}, which means: \begin{aligned}&\left(\frac{n}{n}\right)^n+\dots+\left(\frac{2}{n}\right)^n+\left(\frac{1}{n}\right)^n\\ \geq & \sum_{i=0}^{\left\lfloor n^{1/4} \right\rfloor}\left(1-\frac{i}{n}\right)^n\\ \geq & \sum_{i=0}^{\left\lfloor n^{1/4} \right\rfloor} \left(e^{-i} -\frac{i^2}{n}\right)\\ =& \frac{1-e^{-\left\lfloor n^{1/4} \right\rfloor-1}}{1-e^{-1}}-\lambda \frac{\left\lfloor n^{1/4} \right\rfloor ^3}{n}\end{aligned}
Here \lambda is a constant, and we do not need to know its specific value. Taking the limit, the latter term becomes 0, yielding: \lim_{n\to\infty}\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\dots+\left(\frac{n}{n}\right)^n \geq\frac{e}{e-1}
Combining this with the inequality from the other side, we have: \lim_{n\to\infty}\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\dots+\left(\frac{n}{n}\right)^n = \frac{e}{e-1}
The entire process mainly involves scaling and bounding, and we can scale as we please according to our needs—of course, this is related to the fact that the problem itself is relatively weak. The process described here is not the most elegant, but it is quite practical!
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