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How find this $$\lim_{n\to\infty}\sum_{i=1}^{n}\left(\dfrac{i}{n}\right)^n$$

I think this answer is $\dfrac{e}{e-1}$

and I think this problem have more nice methods,Thank you

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For each fixed $x$, Bernoulli's inequality

$$ (1 + h)^{\alpha} \geq 1 + \alpha h,$$

which holds for $h \geq -1$ and $\alpha \geq 1$, shows that whenever $x \leq n+1$ we have

$$ \left( 1 - \frac{x}{n+1} \right)^{n+1} = \left\{ \left( 1 - \frac{x}{n+1} \right)^{(n+1)/n} \right\}^{n} \geq \left( 1 - \frac{x}{n} \right)^{n}. $$

In particular, it follows that

$$ \left( 1 - \tfrac{k}{n} \right)_{+}^{n} \nearrow e^{-k} \quad \text{as} \quad n\to\infty. $$

Here, the symbol $(x)_{+} := \max\{0, x\}$ denotes the positive part function. Thus whenever $m < n$, the following inequality holds

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

We fix $m$ and take limits to obtain

$$ \liminf_{n\to\infty} \sum_{k=0}^{m} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \liminf_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \limsup_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \sum_{k=0}^{\infty} e^{-k}. \tag{1} $$

But since

$$ \liminf_{n\to\infty} \sum_{k=0}^{m} \left( 1 - \tfrac{k}{n} \right)^{n} = \sum_{k=0}^{m} e^{-k}. $$

taking $m \to \infty$ to the inequality $\text{(1)}$, we have

$$ \sum_{k=0}^{\infty} e^{-k} \leq \liminf_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \limsup_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n} \leq \sum_{k=0}^{\infty} e^{-k}, $$

from which it follows that

$$ \lim_{n\to\infty} \sum_{i=1}^{n} \left( \frac{i}{n} \right)^{n} = \lim_{n\to\infty} \sum_{k=0}^{n-1} \left( 1 - \tfrac{k}{n} \right)^{n} = \sum_{k=0}^{\infty} e^{-k} = \frac{e}{e-1}. $$

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