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I have next problem: $$ e^{H_n} = ... + O(\frac1n)$$

$H_n = \sum_{k=1}^n \frac1k $

I have got the following result:

From ($ H_n > \ln n + \gamma $) I go to $e^{H_n} \geq n e^{\gamma}$

Therefore $e^{H_n} = n e^{\gamma} + ... + O(\frac1n)$

What is the next step to get remaining elements of the sum?

J.Exactor
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  • hmmm why not just put in more terms of the asymptotic expansion of $H_n$ – tired Oct 15 '15 at 22:06
  • You need a more exact approximation for $H_n$ if you want to determine $e^{H_n}$ to the $O(1/n)$ term. Do you know a more exact approximation? – Daniel Fischer Oct 15 '15 at 22:07
  • WolframAlpha says that next element is $\frac{e^{\gamma}}{2}$. But I don't understand how to get this result. – J.Exactor Oct 15 '15 at 22:07
  • Yes, I know $H_n = \ln n + \gamma + \frac1{2n} + O(\frac{1}{n^2})$ – J.Exactor Oct 15 '15 at 22:09
  • I am not sure it is correctly write following:

    $ e^{H_n} = n e^{\gamma + \frac{1}{2n}} e^{O(\frac{1}{n^2})} = n e^{\gamma + \frac{1}{2n}} (1 + O(\frac{1}{n^2}))$

     – J.Exactor Oct 15 '15 at 22:13
  • That is correct, but it's in my opinion better to write $$e^{H_n} = n e^{\gamma} \exp \biggl(\frac{1}{2n} + O(n^{-2})\biggr)$$ and then expand $\exp \bigl(\frac{1}{2n} + O(n^{-2})\bigr) = 1 + \frac{1}{2n} + O(n^{-2})$. – Daniel Fischer Oct 15 '15 at 22:17

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