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

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.

• I don't understand. This is almost like saying $S_n$ is asymptotic to itself. This statement is equivalent to saying that $n/S_n$ vanishes as $n\to\infty$ which is not exactly a sharp asymptotic equivalence, it is just an estimate $S_n\in\omega(n)$. Surely, you would want a better estimate in terms of something more tangible; Stirling's formula is useful, but $n!\sim n!+n$ is not. Oct 12, 2023 at 12:30
• @FShrike, I agree, it looks like a paradox. I guess the point is that in the sum we sum up all $n$ terms, but in the asymptotics we are allowed to keep less than $n$ terms. You may also evaluate $\lim_{n\to\infty}\frac1{\sqrt n}(S_n-2n)=1$ Oct 12, 2023 at 12:47
• I'm just not sure what the question is. About the "allowed to retain" etc. As written this seems like a useless asymptotic estimate for $S_n$ Oct 12, 2023 at 12:57
• You can get $n^{-1/k}((S_n-n)-(n+n^{1/2}+\dots+n^{1/(k-1)}))\to 1$ for all $k\geq 2$ with similar methods. I don't know what your point is, because the extra $n$ comes from all remaining $n^{1/(k+1)},n^{1/(k+2)},\dots, n^{1/n}$ as $\geq 1$ and an estimate of their upper bound. Oct 12, 2023 at 13:37
• @FShrike, I would like to find $k(n)$, so that $$S_n=n+\sum_{i=1}^{k(n)}n^{1/i}+R(n)$$, where the remainder $R(n)=o\big(n^{1/k}\big)$ and, if possible, to get an explicite form of $R(n)$ Oct 12, 2023 at 13:42

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:

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

• (+1) In $O\left(\sum_{k=N+1}^n{\log^2n\over k}\right)$ you should divide by $k^2$.
– Gary
Oct 14, 2023 at 1:12
• Thank you for your solution! I got the similar result (the remainder $O(\ln^2n)$), but in a more complicated way. I think your solution is the shortcut to the answer. Interestingly, we are allowed to retain a very slow growing number ($[\ln n]$) of terms in the asymptotics. Oct 14, 2023 at 5:48
• @Gary Just edited. Thanks for pointing that out Oct 14, 2023 at 14:22
• @Svyatoslav If you are happy to take more terms then you can gain a better error term: $$\sum\limits_{k = 1}^n {n^{1/k} } = 2n + \bigg( {\sum\limits_{k = 2}^{\left\lfloor {\log ^2 (n)} \right\rfloor } {n^{1/k} } } \bigg) - 2\log (n)\log \log (n) + \mathcal{O}(1).$$
– Gary
Oct 16, 2023 at 4:12

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 $$2n + \sqrt n + O{(n^{1/3})}$$.