Find a closed form for $\sum_{k=1}^{x-1} a^{1/k}$ Please find a closed form for partial sum of a function
$$f(x)=a^{1/x}$$
I want it to be expressed in terms of bounded number of elementary functions and/or well known special functions.
No computer algebra systems I have tried so far could find a satisfactory solution. I believe that the expression can exist in the terms of incomplete Gamma function or its generalizations because indefinite integral  of this function can be expressed in terms of incomplete Gamma function:
$$\int f(x) dx= x\sqrt[x]{a}-\operatorname{Ei}\left(\frac{\ln a}{x}\right)\ln a$$

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.
