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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.

Thanks in advance.

Additional Remark

I'm also interested in extending the answer to all $p,q\in \Bbb R^+$.

Edit

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?

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  • $\begingroup$ $H_n = \log(n)+\gamma+o(1)$ as $n\to +\infty$. $\endgroup$ – Jack D'Aurizio Mar 27 at 21:57
  • $\begingroup$ @JackD'Aurizio Thank you for the relation but is there an alternative using the definition of $H_n$? $\endgroup$ – Mostafa Ayaz Mar 27 at 22:02
  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Mar 27 at 22:15
  • $\begingroup$ Thank you so much! $\endgroup$ – Mostafa Ayaz Mar 27 at 22:18

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