Don't limit yourself

First we observe that $\begin{aligned} \sum_{k=0}^{n} \frac{ { n \choose k}}{(k+1)(k+2)} &= \sum_{k=0}^{n} \frac{ {n \choose k}}{(k+1)} - \sum_{k=0}^{n} \frac{ {n \choose k}}{(k+2)} \\ &= \sum_{k=0}^{n}{n \choose k} \cdot \int_{0}^{1} x^{k} \ dx - \sum_{k=0}^{n} {n \choose k} \int_{0}^{1} x^{k+1} \ dx \\ &= \int_{0}^{1} \sum_{k=0}^{n}{n\choose k} x^{k} \ dx -\int_{0}^{1}\sum_{k=0}^{n} {n\choose k} x^{k+1} \ dx \\ &= \int_{0}^{1} (1+x)^{n} \ dx - \int_{0}^{1} x(1+x)^{n} \ dx \\ &= \frac{2^{n+1}-1}{n+1} - \frac{2^{n+1}\cdot n +1}{(n+1)(n+2)} \end{aligned}$

From here on it's easy. Just simply and multiply by $\frac{n^{2}}{2^{n}}$ and we get the answer as $4$

Simply multiply and divide by (n+1)(n+2) as shown below: $L.H.S. = \dfrac {n^2}{2^n(n+1)(n+2)}\sum_{k=0}^{n}\dfrac{{n\choose k}(n+1)(n+2)}{(k+1)(k+2)}$ $= \dfrac {n^2}{2^n(n+1)(n+2)}\sum_{k=0}^{n} {{n+2}\choose {k+2}}$ $=\dfrac{n^2}{2^n(n+1)(n+2)}\cdot[2^{n+2} - 1 - (n-2)]$ Now in the limit, the answer is 4.