Estimation of Summations

$\frac{1}{n} \sum^n_{k=1} n^{1/k}$

By plotting, you can find that the $\frac{1}{n} \sum^\infty_{k=1} k^{1/k}$ finnally goes to $1$, because $\{k^{1/k}\} \to 1$, from which the average goes to $1$ by Cesaro-Stolz lemma.
Now replacing this sum by $\frac{1}{n} \sum^\infty_{k=1} n^{1/k}$, it helps to estimate the new summation by splitting the sum into two pieces.

The summation is

$$n + n^{1/2} + n^{1/3} + \ldots n^{1/n}.$$ The first $\ln(n)$ terms can be estimated by summing $\sqrt n$.

Why would the upper bound $n^{1/k} \leq \sqrt n$ for the first $\ln(n)$ work?? The reason is that we know the sum $\sum_{2 \leq k < \ln n} n^{1/k} \leq \ln(n) \sqrt n = o(n)$, and so this upper bound is sufficient.

Now we estimate $\sum_{\ln n \leq k \leq n} n^{1/k}$.

Inequality

The first inequality to use is $$e^{\ln n /k} \leq \frac{\ln n}{k} + 1$$ for $k\geq \ln n$. ($e^x < 3 \implies e^x < 3x +1$ for $x\in [0,1]$)

Also, for $k \geq 2$,$$1/k = \int^{k}_{k-1} \frac{1}{k} dt <\int^{k}_{k-1} \frac{1}{t} dt$$

So $$\sum^n_{k=2} 1/k < \sum^n_{k=2} \int^k_{k-1} \frac{1}{t}dt = \int^n_{1} \frac{1}{t}dt =\ln n \tag{1}$$

Concluding Estimate

$$\sum_{\ln n < k \leq n} n^{1/k} = \sum_{\ln n < k \leq n} e^{\ln n/k} < 3 \ln n \left(\sum^n_{k=2} \frac{1}{k}\right) + n< 3\ln n \ln n + n = o(n) + n$$