# On the limit of $H_n^p\over n^q$ and $H_{n^p}\over n^q$

As you know, $$H_n$$ is the famous Harmonic number defined as follows:$$H_n=\sum_{k=1}^{n}{1\over k}$$I was wondering for which $$p,q\in \Bbb N$$ do the following limits exist and are equal to some positive real numbers?$$\large\lim_{n\to \infty}{(H_n)^p\over n^q}$$and $$\large\lim_{n\to \infty}{H_{n^p}\over n^q}$$

My attempt for the second one

I tried$${H_{n^p}\over n^q}{={1\over n^q}\sum _{k=1}^{n^p}{1\over k}\\={1\over n^{p+q}}\sum _{k=1}^{n^p}{n^p\over k}\\\sim {1\over n^q}\int_0^1{dx\over x}\\=!!}$$

I finally ended up with an ambiguity $$\infty\over \infty$$ from which I don't know how to get rid of.

Also I don't know how to start solving the first one.

I'm also interested in extending the answer to all $$p,q\in \Bbb R^+$$.
Thanks to Jack D'Aurizio's comment below my question, the relation $$H_n=\ln n+\gamma+o(1)$$ worked out for me, but is there any other proof based on the definition of $$H_n$$? Since $$H_n=\ln n+\gamma+o(1)$$ itself requires a proof, can I ignore it using another way?
• $H_n = \log(n)+\gamma+o(1)$ as $n\to +\infty$. – Jack D'Aurizio Mar 27 at 21:57
• @JackD'Aurizio Thank you for the relation but is there an alternative using the definition of $H_n$? – Mostafa Ayaz Mar 27 at 22:02
• You may simply use that $H_n$ is increasing but $H_{2n}-H_n$ converges to a finite limit, hence $H_n$ grows slower than any polynomial. – Jack D'Aurizio Mar 27 at 22:15