Is there any way to evaluate $e^{H_n} = … + O(\frac{1}{n})$, where $H_n$ is $n$-th harmonic number?

Question

I know, that $H_n = \log n + \gamma + O(1)$, but in that case $e^{H_n} = e^{\log n + \gamma + O(1)} = n e^\gamma e^{O(1)}$ - I can't use this. How can I get this $O(\frac1n)$?

Answer 1

You need another term. $$H_n = \ln(n) + \gamma + \dfrac{1}{2n} + O\left( \dfrac{1}{n^2}\right)$$