any general formula for $a^{1/2} + a^{1/3} + a^{1/4} + a^{1/5} + \cdots+ a^{1/n}$ ?? I met a question like: (if $a$ is an integer)
any general formula for
$$a^{1/2} + a^{1/3} + a^{1/4} + a^{1/5} + \cdots+ a^{1/n}\text{ ??} $$
if a∈N,  how to prove
$$a^{1/2} + a^{1/3} + a^{1/4} + a^{1/5} + +\cdots+ a^{1/n} < a^{2/3}\text{ ??} $$
or how to prove
$$a^{1/2} + a^{1/3} + a^{1/4} + a^{1/5} + +\cdots+ a^{1/n} < a^{K/L}\text{ ??} $$
where K<L; a, K and L are all integers.
so what are K and L?
Thanks
 A: So, first note (as mentioned in multiple comments) that $a^{1/n}\rightarrow 1$ for large $n$; so the partial sums $a^{1/2}+a^{1/3}+\ldots+a^{1/n}$ are at the very least going to be asymptotic to $n$ for large $n$.  (That is, they certainly aren't bounded by any function of $a$.). But you can get a good characterization of the behavior from the following expansion:
$$
\sum_{k=2}^{n}a^{1/k}=\sum_{k=2}^{n}\sum_{i=0}^{\infty}\frac{(\log a)^i}{k^i i!}=\sum_{k=2}^{n}\left(1+\frac{\log a}{k}+\sum_{i=2}^{\infty}\frac{(\log a)^i}{k^i i!}\right)\\ =(n-1)+\left(H_n - 1\right)\log a+\sum_{i=2}^{n}\frac{(\log a)^i}{i!}\sum_{k=2}^{n}\frac{1}{k^i}.
$$
The first term is $\Theta(n)$; the second is $\Theta(\log a \cdot \log n)$; and the remaining terms converge to an analytic function of $\log a$ as $n\rightarrow\infty$.  In particular, we have
$$
\lim_{n\rightarrow\infty}\sum_{i=2}^{n}\frac{(\log a)^i}{i!}\sum_{k=2}^{n}\frac{1}{k^i} =\sum_{i=2}^{\infty}\frac{\zeta(i)-1}{i!}(\log a)^i.
$$
The coefficients go to zero at least as fast as $1/(i!)$, so this is an entire function.
A: If $a>1$ is a real number, then $a^x = f(x)$ is increasing and
$$\begin{align} \sum^{n}_{i=2} a^{1/i} =  & a^{1/2} + a^{1/3} + a^{1/4} + a^{1/5} +\cdots + a^{1/n} \\ 
\leq & (n-1)a^{1/2} \\ 
= & (n - 1) \ln(e^{a^{1/2}}) \\
\leq & e^{a^{1/2} (n- 1)} - 1
\end{align}$$ That was the better I could do for now.
