I am interested in the value of the following sum:
$$\sum_{j=2}^{\sqrt{n}}n^{1/j}$$
where $n$ is an arbitrary positive integer.
It is quite obvious that the value of the sum is less than $n$.
Does anyone know the exact value of this sum? If not, I need a good lower bound, something like $c \cdot n$ if possible (for a constant $c$).
Approximating the sum by an integral, you have
$$\sum_{j=2}^{\lfloor\sqrt n\rfloor}n^{1/j}\approx \sqrt n\,n^{1/\sqrt n}-\log(n)\,\text{Ei}\left(\frac{\log n}{\sqrt n}\right)+C$$
Notice that the first term is $\sqrt n$, and the next ones tend to $1$ pretty slowly. There are $\sqrt n$ of them, which explains that the behavior is above $\sqrt n$.