I am dealing with this limit problem. Let $$U_n=n^{-\frac{1}{2}(1+\frac{1}{n})}\ \text{and}\ V_n=\bigg(\prod_{k=1}^{n}k^k\bigg)^\frac{1}{n^2},\ n>0.$$ Compute $$\lim\limits_{n\to +\infty}U_nV_n.$$ I did it using two different methods and I got different values. I tried to see where was my mistake but I couldn't. \begin{eqnarray*} \ln(U_nV_n)&=& ln U_n+\ln V_n\\ &=& -\dfrac{1}{2}(1+\dfrac{1}{n})\ln n +\dfrac{\ln(1^1)+\ln (2^2)+\cdots +\ln(n^n)}{n^2}\\ &=& -\dfrac{1}{2}\ln n-\dfrac{\ln n}{2n}+\dfrac{\ln 1}{n^2}+\dfrac{2\ln 2}{n^2}+\cdots+\dfrac{n\ln n}{n^2} \end{eqnarray*} From that, we get $\lim\limits_{n\to \infty}\ln (U_nV_n)=-\infty$, which implies that $\lim\limits_{n\to\infty}U_nV_n=0$.
In the other hand, \begin{eqnarray*} \ln(U_nV_n) &=& \ln U_n+\ln V_n\\ &=& -\dfrac{1}{2}(1+\dfrac{1}{n})\ln n+\sum_{k=1}^{n}\dfrac{k\ln k}{n^2}\\ &=& -\dfrac{n+1}{2n}\ln n+\sum_{k=1}^{n}\dfrac{k\ln k}{n^2}\\ &=& -\dfrac{n(n+1)}{2}\cdot \dfrac{1}{n^2}\ln n+\sum_{k=1}^{n}\dfrac{k\ln k}{n^2}\\ &=& \sum_{k=1}^n k\cdot \dfrac{-\ln n}{n^2}+\sum_{k=1}^{n}\dfrac{k\ln k}{n^2}\\ &=& \sum_{k=1}^n\dfrac{k}{n^2}(\ln k-\ln n)=\sum_{k=1}^n\dfrac{k}{n^2}\ln (\dfrac{k}{n})\\ &=& \dfrac{1}{n}\sum_{k=1}^n\dfrac{k}{n}\ln (\dfrac{k}{n})\\ \end{eqnarray*} From that, we get that $\lim\limits_{n\to \infty}\ln (U_nV_n)=\int_0^1x\ln xdx=-\dfrac{1}{4}$, which implies that $\lim\limits_{n\to\infty}U_nV_n=e^{-\frac{1}{4}}$.
Can anyone tell me where is the mistake?
