Showing $ \sum_{k=1}^{n} k^{1/k}\sim n$

This results from two factoids:

Factoid 1: If $f(x)\to\ell$ when $x\to+\infty$, then $\int\limits_0^xf(t)\mathrm dt=\ell\cdot x+o(x)$ when $x\to+\infty$ and $\sum\limits_{k=1}^nf(k)=\ell\cdot n+o(n)$ when $n\to+\infty$.

Factoid 2: $x^{1/x}\to1$ when $x\to+\infty$.

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Edit: The proof of Factoid 1 is the usual boring epsilon-delta stuff. Here we go: for every positive $\varepsilon$, there exists a finite $K_\varepsilon$ such that $|f(x)-\ell|\leqslant\varepsilon$ for every $x\geqslant K_\varepsilon$. Hence, for every $x\geqslant K_\varepsilon$, $$ \left|\int_0^xf(t)\mathrm dt-\ell\cdot x\right|\leqslant C_\varepsilon+\varepsilon x,\quad\text{where}\quad C_\varepsilon=\left|\int_0^{K_\varepsilon}f(t)\mathrm dt-\ell\cdot K_\varepsilon\right|. $$ In particular, $$ \limsup_{x\to+\infty}\left|\frac1x\int_0^xf(t)\mathrm dt-\ell\right|\leqslant\varepsilon, $$ for every $\varepsilon$, that is, $$ \lim_{x\to+\infty}\frac1x\int_0^xf(t)\mathrm dt=\ell. $$ And for the series: : for every positive $\varepsilon$, there exists a finite $N_\varepsilon$ such that $|f(k)-\ell|\leqslant\varepsilon$ for every $k\geqslant N_\varepsilon$. Hence...

An elementary proof:

For $n \ge 4$ we have that

$$ \frac{1 + 1 + \dots + 1 + n^{1/3} + n^{1/3} + n^{1/3}}{n} \ge n^{1/n}$$

using $\text{AM} \ge \text{GM}$ on $n-3$ copies of $1$ and three copies of $n^{1/3}$.

i.e we get the estimate

$$ 1 - \frac{3}{n} + \frac{3}{n^{2/3}} \ge n^{1/n}$$

(From my answer here: How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$?)

And so

$$ n \le \sum_{k=1}^{n} k^{1/k} \le C+ \sum_{k=4}^n (1 - \frac{3}{k} + \frac{3}{k^{2/3}})$$

using $\sum_{k=1}^{n} k^{-2/3} = \mathcal{O}(\int_{1}^{n} x^{-2/3} dx)$ and $\sum_{k=1}^{n} \frac{1}{k} = \mathcal{O}(\log n)$, we get

$$ n \le \sum_{k=1}^{n} k^{1/k} \le n + \mathcal{O}(n^{1/3})$$

$$ 1 \le \frac{1}{n}\sum_{k=1}^{n} k^{1/k} \le 1 + \mathcal{O}(n^{-2/3})$$

And we are done.