Dealing with $e^{n/4}n^{-(n+1)/2}(1^1 2^2...n^n)^{1/n}$

Question

Prove that \begin{align}e^{n/4}n^{-(n+1)/2}(1^1 2^2...n^n)^{1/n}\end{align}

converges to $1$ as $n\to\infty$.

Hint: The logarithm of the given sequence can be written as \begin{align}\sum_k^{n}f\left(\frac{k+\theta} n\right)-n\int_0^1 f(x)\,dx\end{align}

where $f(x)=x\log x$.

I have tried to write it but I was not successful. I got: $\frac{n}{4}-\frac{n+1}{2}\log n+\frac{1}{n}\sum_{k=1}^{n}\log k$ However $\int_0^1 x\log x=\frac{x^2(2\log (x)-1)}{4}|_0^1$

Question:

How can I simplify the expression into $\sum_k^{n}f\left(\frac{k+\theta} n\right)-n\int_0^1 f(x)\,dx$?

Thanks in advance!

Answer 1

Let $$a_n=e^{n/4}n^{-(n+1)/2}(1^1 2^2...n^n)^{1/n}$$we show that $$\lim_{n\to\infty}\ln a_n=0$$Proof: We know that $$\ln a_n=\dfrac{n}{4}-\dfrac{n+1}{2}\ln n-\dfrac{1}{n}\sum_{k=1}^{n}k\ln k=\dfrac{n}{4}-\dfrac{n+1}{2}\ln n-\dfrac{1}{n}\sum_{k=0}^{n}k\ln k$$also $$\dfrac{1}{n}\sum_{k=0}^{n}k\ln k=n\left(\dfrac{1}{n}\sum_{k=0}^{n}\left[\dfrac{k}{n}\ln\dfrac{k}{n}+\dfrac{k}{n}\ln n\right] \right)=n\left(\dfrac{1}{n}\sum_{k=0}^{n}\dfrac{k}{n}\ln\dfrac{k}{n}+\dfrac{n+1}{2n}\ln n \right)$$since $\int_{0}^{1}x\ln xdx=-\dfrac{1}{4}$ as $n$ tends to $\infty$ we may write$$\dfrac{1}{n}\sum_{k=0}^{n}k\ln k\sim n\left(\int_{0}^{1}x\ln xdx+\dfrac{n+1}{2n}\ln n\right)=-\dfrac{1}{4}n+\dfrac{n+1}{2}\ln n$$by substitution we obtain $$\ln a_n\sim 0$$therefore $$\lim_{n\to \infty} a_n=1$$

Answer 2

Without the hint.

Considering $$a_n=e^{\frac n4}n^{-(n+1)/2}\left(\prod_{k=1}^nk^k\right)^{\frac 1 n}$$ taking logarithms as you did $$\log(a_n)=\frac n 4-\frac {n+1}2 \log(n)+\frac 1n \log\left(\prod_{k=1}^nk^k\right)$$and $$\prod_{k=1}^nk^k=H(n)$$ where $H(n)$ is the hyperfactorial function.

Using Stirling like expansions $$\log(H(n))=n^2 \left(\frac{1}{2} \log \left({n}\right)-\frac{1}{4}\right)+\frac{1}{2} n \log \left({n}\right)+\left(\log (A)+\frac{1}{12} \log \left({n}\right)\right)+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right)$$ where $A$ is Glaisher constant $(\approx 1.28243)$.

Then $$\log(a_n)=\frac{12\log (A)+ \log \left({n}\right)}{12n}+\frac{1}{720 n^3}+O\left(\frac{1}{n^5}\right)$$ Continuing with series $$a_n=e^{\log(a_n)}=1+\frac{12\log (A)+ \log \left({n}\right)}{12n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

This also gives a good approximation of $a_n$. For example $$a_{10}=\frac{9\ 21^{7/10} e^{5/2}}{625 \sqrt{2}}\approx 1.04505$$ while the above approximation would give $$1+\frac{\log (A)}{10}+\frac{\log (10)}{120}\approx 1.04406$$