Closed form of $\sum\limits_{i=1}^n k^{1/i}$ or asymptotic equivalent when $n\to\infty$

Cesaro (particular case of the Stolz-Cesaro theorem) says $$ \lim_{n\rightarrow +\infty}a_n=a\quad\Rightarrow \quad \lim_{n\rightarrow+\infty}\frac{\sum_{i=1}^na_i}{n}=a . $$ I assume $k>0$ and $k\neq 1$ (as $S_n=n$ is trivial if $k=1$). Since $\lim_{n\rightarrow +\infty}k^\frac{1}{n}=1$, we get $$ \lim_{n\rightarrow+\infty}\frac{\sum_{i=1}^nk^\frac{1}{i}}{n}=1\quad\Rightarrow\quad S_n=\sum_{i=1}^nk^\frac{1}{i}\;\;\sim\; n. $$ I doubt there is a closed form. But we can go further in the asymptotics. Recall that if $a_n\sim b_n$ and if $b_n $ is (eventually) positive with $\sum_{i=1}^{+\infty}b_i$ divergent, then $\sum_{i=1}^{n}a_i\sim \sum_{i=1}^nb_i$. That's again Stolz-Cesaro if you want. Now by first derivative of $k^x$ at $0$ $$ k^\frac{1}{n}-1\sim\frac{\ln k}{n} \quad\Rightarrow\quad S_n-n=\sum_{i=1}^nk^\frac{1}{i}-1\sim\sum_{i=1}^n\frac{\ln k}{i}=\ln k\cdot H_n\sim\ln k\cdot \ln n $$ where $H_n$ is the $n$th harmonic number. Hence $$ S_n=n+\ln k\cdot\ln n+o(\ln n). $$

Going up to the second derivative of $k^x$, we get $$ k^\frac{1}{n}-1-\frac{\ln k}{n}\sim \frac{(\ln k)^2}{2n^2}. $$ So the resulting series converges and therefore, using $H_n=\ln n +O(1)$,

$$ S_n-n-\ln k \cdot H_n=O(1)\quad\Rightarrow\quad S_n=n+\ln k\cdot \ln n+O(1). $$

Just note that, for $k>1$

$$ \displaystyle S(n,k)=\sum_{i=1}^n k^{1/i} \sim \int_{1}^{n}k^{1/x}dx$$

$$=-\ln(k)\Gamma(0, -\ln(k))-k+\ln(k)\Gamma\left(0, -\frac{\ln(k)}{n}\right)+nk^{1/n} \longrightarrow (1),$$

where $\Gamma(s,x)$ is the upper incomplete gamma function. For the other case $k<1$

$$ \displaystyle S(n,k)=\sum_{i=1}^n k^{1/i} \sim \int_{0}^{n}k^{1/x}dx= n{k}^{1/{n}}+\ln \left( k \right) \Gamma \left( 0,-{\frac{\ln\left( k \right) }{n}} \right) \longrightarrow (2).$$

Here is a numerical example for $k=0.2$ and $n=2000$ the sum is

$$ 1988.361173. $$

and the approximation using $(2)$ is

$$1987.851634. $$