0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ hmmm why not just put in more terms of the asymptotic expansion of $H_n$ $\endgroup$ – tired Oct 15 '15 at 22:06
  • $\begingroup$ 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? $\endgroup$ – Daniel Fischer Oct 15 '15 at 22:07
  • $\begingroup$ WolframAlpha says that next element is $\frac{e^{\gamma}}{2}$. But I don't understand how to get this result. $\endgroup$ – J.Exactor Oct 15 '15 at 22:07
  • $\begingroup$ Yes, I know $H_n = \ln n + \gamma + \frac1{2n} + O(\frac{1}{n^2})$ $\endgroup$ – J.Exactor Oct 15 '15 at 22:09
  • $\begingroup$ 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}))$ $\endgroup$ – J.Exactor Oct 15 '15 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.