8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Hint: substitute $S(n)$ with integral $$\int\limits_0^n m^{-x}dx$$ $\endgroup$ – Norbert May 13 '13 at 10:56
  • $\begingroup$ Ah, of course. Though you mean $m^{x^{-1}}$, right? $\endgroup$ – Thomas Ahle May 13 '13 at 11:23
  • $\begingroup$ yes you are right $\endgroup$ – Norbert May 13 '13 at 11:40
4
$\begingroup$

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 $C$.

$\endgroup$
  • 2
    $\begingroup$ One small correction: $$\sum_{k=1}^n\frac1{k^2}=\frac{\pi^2}{6}\color{#C00000}{-}\frac1n+O(1/n^2)$$ $\endgroup$ – robjohn May 13 '13 at 12:42
  • $\begingroup$ Constante? Habla español? $\endgroup$ – Pedro Tamaroff May 13 '13 at 12:59
  • $\begingroup$ 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 $M$. $\endgroup$ – Thomas Ahle Sep 20 '18 at 22:19
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

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.

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.