Sum of kth roots ($\sum\sqrt[k]{m}$)

I'm trying to find an asymptotic to $$S(n) = \sum_{k=1}^n\sqrt[k]{m}$$ From computational tests, it seems to grow nearly as slowly as $n$. However even $$\sum_{k=1}^\infty\sqrt[k]{m}-1$$ diverges (for $m\neq1$) by the comparison test.

I'm thinking it might be something like $n\log{\log n}$, but I don't know how to show it.

Update: So it turns out to be closer to $n\sqrt[n]{m}$. Does anybody know if there is a nice formula?

• Hint: substitute $S(n)$ with integral $$\int\limits_0^n m^{-x}dx$$ – Norbert May 13 '13 at 10:56
• Ah, of course. Though you mean $m^{x^{-1}}$, right? – Thomas Ahle May 13 '13 at 11:23
• yes you are right – Norbert May 13 '13 at 11:40

You have : $$M^{1/k} = 1 + \frac{1}{k} \ln M + \frac{1}{2} \ln^2 M \frac{1}{k^2} + O(1/k^3)$$ Since $$\sum_{k=1}^n \frac{1}{k} = \ln n + \gamma + \frac{1}{2n} + O(1/n^2)$$ $$\sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6} - \frac{1}{n} + O(1/n^2)$$ we deduce : $$\sum_{k=1}^n M^{1/k} = n + \ln M. \ln n + C + (\ln M + \ln^2 M) \frac{1}{2n} + O(1/n^2),$$ for some constant $C$.

• One small correction: $$\sum_{k=1}^n\frac1{k^2}=\frac{\pi^2}{6}\color{#C00000}{-}\frac1n+O(1/n^2)$$ – robjohn May 13 '13 at 12:42
• Constante? Habla español? – Pedro Tamaroff May 13 '13 at 12:59
• This is very nice in terms of $n$, but of course, it doesn't capture growth in terms of $M$, since the $O(1/n^2)$ term contains arbitrarily large $\Omega(\ln^k M)$ terms; and we know the sum grows at least as fast as $M$. – Thomas Ahle Sep 20 '18 at 22:19

For small enough $x>0$ we have $e^x-1\in(x,2x)$. Furthermore, $\sqrt[k]{m}-1=m^{1/k}-1=e^{(\ln m)/k}-1$. As $k$ increases $x$ will approach $0$ and hence for large enough $k$ we have $$\frac{\ln m}{k}<\sqrt[k]{m}-1<\frac{2\ln m}{k}\,.$$ Since $$\sum_{k=1}^n\frac1k \sim \ln n \longrightarrow \infty$$ for $n\to\infty$, by the sandwich lemma, your sum will also go to infinity for $n\to\infty$ with logarithmic speed.

After Ralph's answer I'd like to detail this a bit. [update](upps, after I've posted this it seems user10670 was 30 sec's ahead)

After some standard-manipulation with the Carleman-matrices for $$m \to \sqrt[x]{1+m}-1 =\exp( \log(1+m) / x)-1$$
I come to the following asymptotic.
We need also the expression for the sum of the first consecutive reciprocals $$s_p(a,n) = \sum_{k=a}^n \frac 1{k^p}$$ then we get $$S_m(n) = \sum_{k=0}^\infty {\log(m)^k s_k(2,n) \over k!}$$ and clearly this contains a $\zeta(1)-1$-expression in $s_1(2,n)$ if n is assumed to go to infinity.

unchecked remark: To get this possibly converging I think you should not only reduce by 1 but also by the next term, so something like $$S_m(n) = \sum_{k=1}^n \sqrt[k]{m}-1+m/k$$ to eliminate the $\zeta(1)$-expression

• Given $\sum_{k=1}^n\sqrt[k]{m}-1$ diverges against positive infinity, I don't think this will work.. – Thomas Ahle May 14 '13 at 12:48

We can get some intuition for this by taking the integral $$\int_1^n m^{1/x} = n + \frac{m}{\log m} + (\log m) (\log n-\log\log m-\gamma+1) + O\left(\frac{m}{\log^2 m} + \frac{\log^2 m}{n}+\frac{1}{m^2}\right)$$.

However, we know that the actual result has to grow at least as fast as $$m$$.

The best, reasonably simple, formula I've found so far is: $$m + n + \log n \log m - 1$$.

If $$m$$ and $$n$$ grows at the same speed, it doesn't quite grow fast enough (loses about 0.2% at $$m=n=10^5$$), but if one of $$m$$ or $$n$$ grows fastest, it is nearly perfect.

Plotted with $$m=2$$, $$0