Showing $ \sum_{k=1}^{n} k^{1/k}\sim n$ I would like to show that:
$$ \sum_{k=1}^{n} k^{1/k} \sim n$$
by using integrals.
We have:
$$ \int_{3}^{n+1} t^{1/t} \mathrm dt +\sqrt{2}+1 \leq \sum_{k=1}^{n} k^{1/k} \leq \int_{3}^{n} t^{1/t} \mathrm dt +3^{1/3}+\sqrt{2}+1$$
$$ \int_{3}^{n+1} t^{1/t} \mathrm dt= (n+1)^{1+\frac{1}{n+1}}-3^{1/3+1}+\int_{3}^{n+1}(\ln(t)-1)t^{1/t-1}\mathrm dt $$ 
$$ \int_{3}^{n} t^{1/t} \mathrm dt= n^{1+\frac{1}{n}}-3^{1/3+1}+\int_{3}^{n}(\ln(t)-1)t^{1/t-1}\mathrm dt $$ 
So I have to show that:
$$ \int_{3}^{n}(\ln(t)-1)t^{1/t-1}\mathrm dt=o(n) $$
$$ \int_{3}^{n+1}(\ln(t)-1)t^{1/t-1}\mathrm dt=o(n) $$
which is my question.
Edit: I'm now trying to show that: $$ \sum_{k=1}^{n} k^{1/k}=n+\frac{\ln(n)^2}{2}+O(1)$$
We have:
$$ \sum_{k=1}^{n} k^{1/k}=n+\sum_{k=1}^{n} \frac{\ln(k)}{k}+O(1)$$
Using integrals: $$ \sum_{k=1}^{n} \frac{\ln(k)}{k}=\frac{\ln(n)^2}{2}+o(\frac{\ln(n)^2}{2}) $$
However: why $$ \sum_{k=1}^{n} \frac{\ln(k)}{k}-\frac{\ln(n)^2}{2}=O(1) $$ ?
 A: 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 $x\to+\infty$.

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...
A: 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.
A: Let $$ P_0(x) = x -  \lfloor x\rfloor\ - \frac{1}{2} = \left\{x\right\} - \frac{1}{2} $$
You can use Euler-Maclaurin summation:
$$ \int_1^{n} x^{1/x} dx = \sum_{k=1}^{n-1} \int_k^{k+1} x^{1/x} dx = \sum_{k=1}^{n-1} \bigg|_k^{k+1} P_0(x) x^{1/x} - \int_k^{k+1} P_0(x) \ x^{1/x} \ \frac{1- \ln(x)}{x^2} \ dx   $$
$$ = \sum_{k=1}^{n-1} \left( \frac{1}{2} (k+1)^{1/(k+1)} + \frac{1}{2} k^{1/k} \right) + \int_1^{n} P_0(x) \ x^{1/x} \ \frac{\ln(x)-1}{x^2} \ dx $$
$$ = \sum_{k=1}^n k^{1/k} - \frac{1}{2} 1^{1/1} - \frac{1}{2} n^{1/n} +  \int_1^{n} P_0(x) \ x^{1/x} \ \frac{\ln(x)-1}{x^2} \ dx $$
and so $$ \sum_{k=1}^n k^{1/k} = \int_1^n x^{1/x} dx  + \frac{1}{2} + \frac{1}{2} n^{1/n} - \int_1^{n} P_0(x) \ x^{1/x} \ \frac{\ln(x)-1}{x^2} \ dx  $$ Now, $P_0(x)x^{1/x}$ is bounded, and $\frac{\ln(x)-1}{x^2}$ behaves like $\frac{\ln(x)}{x^2}$ for large $x$, whose integral converges, so the right hand integral converges, and $n^{1/n} \to 1$ as $n \to \infty$, so:
$$ \sum_{k=1}^n k^{1/k} = \int_1^n x^{1/x} dx + \left( 1 - \int_1^{\infty} P_0(x) \ x^{1/x} \ \frac{\ln(x)-1}{x^2} \ dx \right) + o(1)   $$
(note the term in the brackets is just a constant), so all that is left is to approximate the first integral:
$$ \int_1^n x^{1/x} dx = \int_1^n dx + \int_1^n \frac{\ln(x)}{x} dx + \int_1^n \left( x^{1/x} - 1 - \frac{\ln(x)}{x} \right) dx  $$
$$ = n - 1 + \frac{\ln(n)^2}{2} + \int_1^n \left( x^{1/x} - 1 - \frac{\ln(x)}{x} \right) dx  $$
A Taylor expansion shows that $$ \left( x^{1/x} - 1 - \frac{\ln(x)}{x} \right) \sim \frac{\ln(x)^2}{2x^2} $$ as $x \to \infty$, and so the final integral converges, so
$$  \int_1^n x^{1/x} dx = n + \frac{\ln(n)^2}{2} -1 + \int_1^{\infty} \left( x^{1/x} - 1 - \frac{\ln(x)}{x} \right) dx   + o(1)$$ 
Putting it all together:
$$ \sum_{k=1}^n k^{1/k} =  n + \frac{\ln(n)^2}{2} +  \int_1^{\infty} \left( x^{1/x} - 1 - \frac{\ln(x)}{x} \right) dx - \int_1^{\infty} P_0(x) \ x^{1/x} \ \frac{\ln(x)-1}{x^2} \ dx +o(1)  $$
i.e. $$\sum_{k=1}^n k^{1/k} =  n + \frac{\ln(n)^2}{2} + c + o(1) $$ for a constant $c$. If you want even $\textit{further}$ terms in the asymptotic expansion, you can use this same method, and simply approximate each of the integrals to further accuracy, for the first integral by using more terms of the taylor expansion, for the second integral by integrating by parts using the Bernoulli polynomials, you can look them up in the Wikipedia page on Euler Maclaurin Summation.
