3
$\begingroup$

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?

  • 0
    Are you looking for the closed form of $a_n$, i.e. it should have a nice expression through the elementary functions?2012-02-22
  • 1
    This should help http://www.wolframalpha.com/input/?i=Integrate%5BLog%5Bk%5D%5E%281%2F3%29%2C+%7Bk%2C+1%2C+n%7D%2C+Assumptions+-%3E+n+%3E+1%5D2012-02-22
  • 1
    $\root3\of{\log k}$ is a very slowly growing function, so the sum should be asymptotic to $n\root3\of{\log n}$.2012-02-22
  • 0
    How about $1-\frac{1}{x} \leq ln(x) \leq x-1$2012-02-22
  • 0
    @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

2 Answers 2

6

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.

  • 0
    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?2012-02-22
  • 0
    @PlaneChon-Ju I think it's an application of L'Hôpital and the fundamental theorem of calculus.2012-02-22
2

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.

  • 0
    $$ \int_{1}^n \ln(t)^{1/3} dt= n\ln(n)^{1/3}-\frac{1}{3}\int_{1}^{n} \frac{1}{\ln(t)^{2/3}}dt $$ But how can I show that $$\int_{1}^{n} \frac{1}{\ln(t)^{2/3}}dt=o(n\ln(n)^{1/3}) $$2012-02-22
  • 0
    @PlaneChon-Ju: Right, but that does not matter for the final thing. I will correct. Thanks.2012-02-22
  • 0
    @PlaneChon-Ju: $(\ln(t))^{2/3} \ge 1$ for $t \ge e$.2012-02-22