Product fractions limit $\lim_{n\to \infty} (\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}})^{1/n^2}$ I need some help with this limit
$$\lim_{n\to \infty} \left(\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}}\right)^{1/n^2}$$
This problem appears in a chapter about Riemann sums, so I think I must split the fractions in the parantheses:
$$\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}} = \frac{1^1}{n^1}\times \frac{2^2}{n^2}\times ... \times \frac{n^n}{n^n}$$
Now I am stuck.
 A: Hint. By taking the logarithm we find
$$\log\left(\left(\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}}\right)^{1/n^2}\right)=\frac{1}{n}\sum_{k=1}^n (k/n)\log(k/n)$$
which looks like a Riemann sum...
A: Let $I$ denote the given limit. Then
$$I = \lim_{n\to\infty}\left(\prod_{i=1}^n\left(\dfrac{i}{n}\right)^i\right)^{\frac{1}{n^2}}$$
$$=e^{\lim_{n\to\infty}\frac{1}{n^2}\sum_{i=1}^n\left(i\log\frac{i}{n}\right)}$$
$$ = e^{\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\left(\frac{i}{n}\log\frac{i}{n}\right)}$$
$$=e^{\int_0^1x\log x\mathrm{d}x}$$
Now apply integration by parts.
A: $$\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)^{\frac{1}{n^{2}}}=\exp\left(\frac{1}{n^{2}}\ln\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)\right)$$$$=\exp\left(\frac{1}{n^{2}}\ln\left(\prod_{k=1}^{n}\left(\frac{k}{n}\right)^{k}\right)\right)=\exp\left(\frac{1}{n}\sum_{k=1}^{n}\ln\left(\left(\frac{k}{n}\right)^{\frac{k}{n}}\right)\right)$$$$=\exp\left(\int_{0}^{1}\ln\left(\left(x\right)^{x}\right)dx\right)=\exp\left(\int_{0}^{1}x\ln\left(\left(x\right)\right)dx\right)$$$$=\exp\left(\frac{x^{2}}{2}\ln\left(\left(x\right)\right)\Big|_0^1-\frac{1}{2}\int_{0}^{1}xdx\right)$$$$=\color{red}{\exp\left(\frac{-1}{4}\right)\simeq0.778800783071}$$
A: $$A=\lim_{n\to\infty}\left(\prod_{r=1}^n\left(\dfrac rn\right)^{r/n}\right)^{1/n}$$
$$\ln A=\lim_{n\to\infty}\dfrac1n f\left(\dfrac rn\right)$$  where $f(x)=x\ln x$ 
Now use The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$  and integrate by parts using LIATE : How does it work?
A: Just for your curiosity
It is clear that Riemann sum make the problem quite simple.
But we can go beyond the limit and approximate the term which is
$$a_n=\left(n^{-\frac{1}{2} n (n+1)} H(n)\right)^{\frac{1}{n^2}}\implies \log(a_n)=\frac 1{n^2}\log\left(n^{-\frac{1}{2} n (n+1)} H(n)\right)$$ where $H(n)$ is the hyperfactorial function.
Expanding the logarithm and using the asymptotics of $\log(H(n))$ we quickly end with
$$\log(a_n)=\frac{1}{ n^2}\left(-\frac{n^2}{4}+\log (A)+\frac{1}{12} \log
   \left({n}\right)+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right)\right)$$ where $A$ is Glaisher constant that is to say
$$\log(a_n)=-\frac{1}{4}+\frac{12\log (A)+ \log
   \left({n}\right)}{n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it ia approached.
Moreover, this gives you a shortcut method for computing $a_n$. If you are patient, compute $a_{10}$; the exact value is $0.78224014$ while the above (very) truncated series would give                           $0.78224004$.
