Just for fun here are some bounds for $\sum_{k=1}^n a^{1/k}$ for $ a \ge 1.$
We have
$a^{1/k} = 1 + \frac{\log a}{1! k} + \frac{(\log a)^2}{2! k^2} + \frac{(\log a)^3}{3! k^3} + \cdots$
and so
$\sum_{k=1}^n a^{1/k} = \zeta_n(0) + (\log a)\zeta_n(1) + \frac{(\log a)^2}{2!}\zeta_n(2) + \frac{(\log a)^3}{3!}\zeta_n(3) + \cdots$
where $\zeta_n(r) = \sum_{k=1}^n 1/k^r.$ Thus $\zeta_n(0)=n$ and $\zeta_n(1)=H_n = 1 + 1/2 + 1/3 + \cdots + 1/n.$
Now $H_n= \log(n + 1/2) + \gamma + \epsilon(n),$ where $0< \epsilon(n)< 1/24n^2$ and $\gamma$ is the Euler-Mascheroni constant, and (by comparing the sum with the integral of $1/x^r$)
$ \frac{1}{(r-1)(n+1)^{r-1}} < \sum_{k=n+1}^\infty \frac{1}{k^r} < \frac{1}{(r-1)n^{r-1}}.$
Hence
$ \frac{1}{(r-1)(n+1)^{r-1}} < \zeta(r) - \zeta_n(r) < \frac{1}{(r-1)n^{r-1}}$
and so we have
$ l(a) \le \sum_{k=1}^n a^{1/k} \le u(a) $
where
$l(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma \right) +\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n} \right)$ $+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2n^2} \right) +\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3n^3} \right) + \cdots$
and
$u(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma + \frac{1}{24n^2} \right) +\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n+1} \right)$ $+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2(n+1)^2} \right) +\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3(n+1)^3} \right) + \cdots.$
Providing $a$ is not enormous and $n$ is sufficiently large, we only need a few terms for a good approximation to our sum.