Is there a closed form for $\sum\limits_{k = 2}^n \sqrt[k] n$ for some natural number $n$? As titled. I'm still thinking about this question. Thanks for any discussions on this! 
 A: (Too long for a comment.) $\;x_n=\sum\limits_{k = 2}^n \sqrt[k] n\,$ is an algebraic number whose minimal polynomial can be explicitly determined, but is unlikely to be solvable in a "closed form" any simpler or more general than the sum that originally defined it. First few cases:


*

*$n=2\,$: $\;x_2=\sqrt{2}\,$ satisfies $\;x^2-2=0$ 

*$n=3\,$: $\;x_3\,$ satisfies $\;x^6 - 9 x^4 - 6 x^3 + 27 x^2 - 54 x - 18=0$

*$n=4\,$: $\;x_4\,$ satisfies $\;x^6 - 12 x^5 + 54 x^4 - 120 x^3 + 156 x^2 - 192 x + 184=0$

*$n=5\,$: $\;x_5\,$ satisfies a $60^{th}$ degree polynomial
A: Not sure whether a closed form exists but using Jensen's inequality we can derive a lower bound as follows.
Let $X$ be a random variable with sample space $\left\{2, 3, \ldots, n \right\}$ and with probability $P(X=i) = \frac{1}{n-1}$. Let $g(x) = n^{\frac{1}{x}}$ which is a convex function when $x>0$. By Jensen's inequality,
$$Eg(X) \geq g(EX)$$
$$\implies \frac{\sum\limits_{k=2}^{n}n^{\frac{1}{k}}}{n-1} \geq g\left(\frac{\sum\limits_{k=2}^{n}k}{n-1}\right) = g\left(\frac{\frac{n(n+1)}{2}-1}{n-1}\right) = g\left(\frac{n+2}{2}\right) = n^{\frac{2}{n+2}}$$
$$\implies \sum\limits_{k=2}^{n}n^{\frac{1}{k}} \geq (n-1)n^{\frac{2}{n+2}}$$
A: No.
But notice
$$\lim_{n \to \infty}\sum_{k=2}^{n} n^{\frac{1}{k}-1} = 1$$
being equal to
$$\lim_{x \to \infty} \int_{2}^x x^{\frac{1}{y}-1} {d}y$$
This second is simpler to evaluate, although even that on its own does not have a closed form without using exponential or similar integral.
