Asymptotics of the sum $\sum_{k=1}^nn^{1/k}$ at $n\to\infty$

Question

The asymptotics of the sum $\displaystyle S_n=\sum_{k=1}^nn^{1/k}$ can be found, for example, by means of Stolz-Césaro theorem (for example, here); I found it via the Euler-Maclaurin' summation formula.

The answer is $$S_n=n+n^{1/2}+n^{1/3}+n^{1/4}+...+n^{1/n}\sim 2n+n^{1/2}+n^{1/3}+n^{1/4}+...$$ As we deal with the asymptotic series, we may expect that there is no contradiction here.

So, my questions are:

1. How many term of the asymptotics (depending on $n$) we are allowed to retain?
2. How we can evaluate (or estimate) the remainder term?

Thank you.

Answer 1

Let $N$ be some positive integer to be determined later, so we have

$$ S_n=\sum_{k=1}^Nn^{1/k}+\sum_{k=N+1}^nn^{1/k}. $$

Notice that when $k>\log n$, we have $n^{1/k}=1+{\log n\over k}+O\left(\log^2n\over k^2\right)$. This indicates that when $N=\lfloor\log n\rfloor$, there is

$$ \begin{aligned} \sum_{k=N+1}^nn^{1/k} &=\sum_{k=N+1}^n1+\sum_{k=N+1}^n{\log n\over k}+O\left(\sum_{k=N+1}^n{\log^2n\over k^2}\right) \\ &=n-N+(\log n)\log{n\over N+1}+O\left(\log n\over N\right)+O\left(\log^2n\over N\right) \\ &=n+O(\log^2n). \end{aligned} $$

Consequently, we have the following asymptotic expansion:

$$ S_n=2n+n^{1/2}+n^{1/3}+\dots+n^{1/\lfloor\log n\rfloor}+O(\log^2n). $$

Answer 2

You add $n$ terms each $\ge 1$, that’s $n$ in total. The first term is $n$, the second $\sqrt n$, the third $n^{1/3}$, I bet the sum is $2n + \sqrt n + O{(n^{1/3})}$.