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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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    @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?2012-02-22

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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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    @PlaneChon-Ju I think it's an application of L'Hôpital and the fundamental theorem of calculus.2012-02-22
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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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    @PlaneChon-Ju: $(\ln(t))^{2/3} \ge 1$ for $t \ge e$.2012-02-22