Evaluate the limit!

$\large \displaystyle L = \lim_{n \rightarrow \infty} \frac{1^n + 2^n + 3^n + ... + n^n}{n^n}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \frac{1^n}{n^n} + \frac{2^n}{n^n} + \frac{3^n}{n^n} + ... + 1$

Reversing the order of the summation, i.e., going from the last to the first term:

$\large \displaystyle L = \lim_{n \rightarrow \infty} 1 + \left( \frac{n-1}{n} \right )^n + \left( \frac{n-2}{n} \right )^n + \left( \frac{n-3}{n} \right )^n + ...$

$\large \displaystyle L = \lim_{n \rightarrow \infty} 1 + \left( 1 - \frac1n \right )^n + \left( 1 -\frac2n \right )^n + \left( 1 - \frac3n \right )^n + ...$

Now, a well know fact that comes from the definition of $e$ is that:

$\large \displaystyle \lim_{n \rightarrow \infty} \left (1 + \frac{a}{n} \right) ^{bn} = e^{ab}$

So:

$\large \displaystyle L = 1 + e^{-1} + e^{-2} + e^{-3} + ...$

$\color{#20A900} \boxed{ \large \displaystyle L = \frac{1}{1-e^{-1}} \approx 1.582}$

Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle \lfloor L \rfloor = 158 }$

I will just work on upper bound( different way) for this problem and for more rigorous work here are details by Sir , Mark Henning .

Note that $\left(\frac{k}{n}\right)^n=\left(1+\frac{n-k}{k}\right)^{-n}\leq e^{-(n-k)}$ so $\sum_{k=1}^n \left(\frac{k}{n}\right)^{n}\leq \sum_{k=1}^{n} e^{k-n}=\sum_{m=1}^{n-1} e^{-m}$ therefore the upper bound we seek is $\lim_{n\to \infty} \operatorname{sup}\sum_{k=1}^n\left(\frac{k}{n}\right)^n\leq \sum_{ m\geq 0} e^{-m}=\frac{e}{e-1}$

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More information : This is just particular case of Wolstenholme limit for the real positive number $m=1$ . That is $\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{mn}=\frac{e^m}{e^m-1}$ set $m=1$ we have our required result as $\frac{e}{e-1}$ .

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The variant version was proposed by J. A Scott (UK) which was published by American mathematical monthly, Vol 126, November 2019 as $\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{k}$ the result is same that of Wolstenholme limit .

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I manage to find(before I discover the fact about Wolstenholme limit) the approximation / inequality for J.A Scott (Uk), which I proposed and published by RMM , here

Show that $\lim_{n\to\infty} \sum_{k=1}^{n}\left(\frac{k}{n}\right)^k < \frac{\ln 3}{\ln 2}$