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How can I find $ a_{n}$ such that $$a_{n} \sim_{n \rightarrow \infty} \sum_{k=1}^n (\ln k)^{1/3} $$ ?

I tried to use integrals:

$$ \int_{k-1}^{k} \ln(t)^{1/3} \mathrm dt\leq \ln(k)^{1/3}\leq \int_{k}^{k+1} \ln(t)^{1/3} \mathrm dt$$ but I cannot compute $$\int_{k-1}^{k} \ln(t)^{1/3} \mathrm dt, \int_{k}^{k+1}\ln(t)^{1/3} \mathrm dt$$

Any idea?

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Are you looking for the closed form of $a_n$, i.e. it should have a nice expression through the elementary functions? –  Ilya Feb 22 '12 at 11:17
This should help wolframalpha.com/input/… –  Norbert Feb 22 '12 at 11:34
$\root3\of{\log k}$ is a very slowly growing function, so the sum should be asymptotic to $n\root3\of{\log n}$. –  Gerry Myerson Feb 22 '12 at 12:12
How about $1-\frac{1}{x} \leq ln(x) \leq x-1$ –  89085731 Feb 22 '12 at 12:24
@Gingerjin, that's not a very sharp estimate when $x$ is large. Have you tried to see whether it's good enough to get an answer to the question? –  Gerry Myerson Feb 22 '12 at 23:23
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2 Answers 2

From your work, it follows that $$ \sum_{k=1}^n (\ln k)^{1/3}\sim\int_{1}^{n} \ln(t)^{1/3} \, dt. $$ Now, for any $p>0$ $$ \lim_{r\to\infty}\frac{\int_{1}^{r}(\ln t)^p\,dt}{r\,(\ln r)^p}=\lim_{r\to\infty}\frac{(\ln r)^p}{(\ln r)^p+p\,r\,(\ln r)^{p-1}\dfrac{~1}{r}}=\lim_{r\to\infty}\frac{1}{1+\dfrac{p}{\ln r}}=1, $$ so that $$ \sum_{k=1}^n (\ln k)^{1/3}\sim n\,(\ln n)^{1/3}. $$ This is precisely the asymptotic behaviour given in Gerry's comment.

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I don't understand the equality: $$\lim_{r\to\infty}\frac{\int_{1}^{r}(\ln t)^p\,dt}{r\,(\ln r)^p}=\lim_{r\to\infty}\frac{(\ln r)^p}{(\ln r)^p+p\,r\,(\ln r)^{p-1}\dfrac{~1}{r}}$$ Could you be more precise? –  Chon Feb 22 '12 at 22:22
@PlaneChon-Ju I think it's an application of L'Hôpital and the fundamental theorem of calculus. –  Dylan Moreland Feb 22 '12 at 22:31
Thank you very much! –  Chon Feb 22 '12 at 22:53
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You don't need to guess $a_n$.

Using integration by parts we get $$\int_{2}^{n} \sqrt[3]{\log x} dx = n \sqrt[3]{\log n} - C - \frac{1}{3} \int_{2}^{n} (\log x)^{-2/3} dx = n \sqrt[3]{\log n}+ \mathcal{O}(n)$$

When powers of the $\log$ function are involved (say $(\log x)^\alpha$), it is usually a good idea to try integration by parts, taking $u = 1$, $v = (\log x)^\alpha$.

See Also: Euler McLaurin Summation formula.

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$$ \int_{1}^n \ln(t)^{1/3} dt= n\ln(n)^{1/3}-\frac{1}{3}\int_{1}^{n} \frac{1}{\ln(t)^{2/3}}dt $$ But how can I show that $$\int_{1}^{n} \frac{1}{\ln(t)^{2/3}}dt=o(n\ln(n)^{1/3}) $$ –  Chon Feb 22 '12 at 22:18
@PlaneChon-Ju: Right, but that does not matter for the final thing. I will correct. Thanks. –  Aryabhata Feb 22 '12 at 22:19
@PlaneChon-Ju: $(\ln(t))^{2/3} \ge 1$ for $t \ge e$. –  Aryabhata Feb 22 '12 at 22:22
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