Compute $\lim_{n \to +\infty} n^{-\frac12 \left(1+\frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\frac{1}{n^2}}$ 
How to compute
  $$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$
  I'm interested in more ways of computing limit for this expression

My proof:
Let $u_n$be that sequence we've:
\begin{eqnarray*}
\ln u_n &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln k\\
&=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}+\frac{1}{n^2}\sum_{k=1}^n k\ln n\\
&=&  \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}\\
&=& \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\ln \frac{k}{n}\\
&\to&\int_0^1 x\ln x\,dx = -1/4
\end{eqnarray*}
Therefore the limit is $e^{-\frac{1}{4}}$

We have $$\exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {n^2} (1 \log 1 + 2 \log 2 + \cdots + n \log n)\right) = \exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {2} \log n - \frac {1} {4} + \frac {\log n} {2n} + o \left (\frac {1} {n}\right) \right) = \exp \left (-\frac {1} {4} + o \left (\frac {1} {n}\right) \right),$$ using the fact that $$1 \log 1 + 2 \log 2 + \cdots + n \log n = \frac {1} {2} n^2 \log n - \frac {1} {4} n^2 + \frac {1} {2} n \log n + o (n).$$ By letting $n \to \infty$, we have $$\exp \left (-\frac {1} {4} + o \left (\frac {1} {n}\right) \right) \to \exp (-\frac {1} {4}).$$
