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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?

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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
up vote 4 down vote accepted

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 $C$.

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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
How do I estimate the constant? – Thomas Ahle May 13 '13 at 13:25
Also, the sum is at least as large as $m$. When $m$ is larger than $n$ the formula seems to miss the target quite badly. Is this somehow hidden in the constant? – Thomas Ahle May 13 '13 at 13:40
I tried using the expansion at $k=1$ instead, but it only gave good results for very small n. – Thomas Ahle May 13 '13 at 13:44

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.

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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

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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

The best, reasonably simple, formula I've found so far is: $$m + n + log(n)log(m) - 1$$ Which is interestingly symmetric.

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.

enter image description here

Plotted with $m=2$, $0<n\leq10$

This is done using user10676's method, but starting from $k=2$ and adding the $k=1$ case manually. I replaced the $H_n-1$ term by $log(n)$ as it was simpler and gave better results.

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