No Technique Needed?

We have that: $\displaystyle \ln\left[\dfrac{1}{2^n}\prod_{k=1}^n \left(2+\dfrac{k}{n^2}\right)\right]=\sum_{k=1}^n \ln\left(1+\dfrac{k}{2n^2}\right)$

and since $x-\dfrac{x^2}{2}<\ln(1+x)<x\qquad\forall x>0$

then $\displaystyle \sum_{k=1}^n \dfrac{k}{2n^2}-\dfrac{1}{2}\sum_{k=1}^n \dfrac{k^2}{4n^4}<\ln\left[\dfrac{1}{2^n}\prod_{k=1}^n \left(2+\dfrac{k}{n^2}\right)\right]<\sum_{k=1}^n \dfrac{k}{2n^2}$ .

But $\displaystyle \sum_{k=1}^n \dfrac{k}{2n^2}=\dfrac{1}{2n^2}\sum_{k=1}^n k=\dfrac{n(n+1)}{4n^2}\longrightarrow\dfrac{1}{4}$

and $\displaystyle \sum_{k=1}^n \dfrac{k^2}{4n^4}=\dfrac{1}{4n^4}\sum_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}{24n^4}\longrightarrow0$

So $\displaystyle \lim_{n\to\infty}\ln\left[\dfrac{1}{2^n}\prod_{k=1}^n \left(2+\dfrac{k}{n^2}\right)\right]=\dfrac{1}{4}=\boxed{0.25}$