Evaluate the following : $\lim_{n \to \infty} \left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}}$ where $H_k = \sum_{i=1}^{k} \frac{1}{i}$.

Question

Evaluate the following limit : $$\lim_{n \to \infty} \left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}}$$ where $$H_k = \sum_{i=1}^{k} \frac{1}{i}.$$

This question was from this; I tried some inequalities with $H_k$ and $\ln k$, but I can't make it further. Is this limit convergent, or divergent? Can we evaluate it if it converges?

Answer 1

By Stirling's approximation, for $n\gg0$, we have $n!\ge (n/e)^n$. Moreover, $\prod_{k=1}^nH_k\le H_n^n$, so that for $n\gg0$, we have:

$$\left(\frac{n!}{\prod_{k=1}^n H_k}\right)^{1/n}\ge\frac{n/e}{H_n}\to\infty.$$

Answer 2

By "ratio to root criterion", we have that

$$\frac{a_{n+1}}{a_n} \to L\in \mathbb R \cup \{\infty\} \implies \sqrt[n] {a_n} \to L $$

and since

$$\frac{(n+1)!}{\prod_{k=1}^{n+1} H_k}\frac{\prod_{k=1}^{n} H_k}{n!}=\frac {n+1} {H_{n+1}} \sim \frac {n}{\log n} \to \infty$$

therefore

$$\lim_{n \to \infty} \left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}} \to \infty$$

Refer to

• Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$

Answer 3

$$\left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}}=\left\{\prod_{k=1}^{n} \frac{k}{H_k} \right\}^{\frac{1}{n}}$$

Now, because $H_k = \log(k) +O(1)$, $\frac{k}{H_k} \to \infty $ and hence the geometric mean also tends to infinity.

Let's do a informal approximation using $H_k \approx \log(k) +\gamma$

Letting $x = k/n$:

$$\begin{align} \log( H_k) &\approx \log( \log n +\log x +\gamma)\\ &=\log \log n+ \log\left(1+ \frac{\log x + \gamma}{\log n}\right)\\ &\approx \log \log n+ \frac{1}{\log n}(\log x + \gamma) \end{align}$$

and

$$ \log \frac{k}{H_k} \approx \log(n) + \log(x) - \log \log n -\frac{1}{\log n}(\log x + \gamma) $$

Hence, because $\int_0^1 \log(x)dx =-1$: $$ \frac{1}{n} \sum_{k=1}^n \log \frac{k}{H_k} \approx \log(n) - \log \log n - 1 + \frac{1-\gamma }{\log n} = \log\left(\frac{n}{\log n }\right) - 1 + o(1) $$

(it indeed tends to infinity, quite slowly) and

$$ \left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}}\approx \frac{n}{ e \log(n)} \to \infty $$

Answer 4

$$a_n=\Bigg[\frac{n!}{\prod_{k=1}^{n} H_k} \Bigg]^{\frac{1}{n}}$$ $$\log(a_n)={\frac{1}{n}}\left(\log(n!)-\log\left(\prod_{k=1}^{n} H_k \right)\right)$$

Looking at @Marty Cohen's answer here $$\log\left(\prod_{k=1}^{n} H_k \right)\simeq n \log (\log (n))+\frac{(\gamma -1) n}{\log (n)}+O\left(\log (\log (n))\right)$$ Using Stirling approximation $$\log(a_n)=\left(\log (n)-\log (\log (n))+\frac{1-\gamma }{\log (n)}-1\right)+\frac{\log (2 \pi n)}{2 n}+\cdots$$ $$a_n=e^{\log(a_n)}\simeq \frac{n}{e \log (n)}\exp\Bigg[\frac{1-\gamma }{\log (n)}+\frac{\log (2 \pi n)}{2 n}+\cdots\Bigg]$$

Trying for $n=1234$, the "exact" result is $68.9624$ while the above gives $67.9256$