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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?

  • 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

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