It is obviosly: $\lim_{x\to 0}\frac{ \displaystyle (1+x^2)^{n+1}-\prod_{k=1}^n\cos kx}{ \displaystyle x\sum_{k=1}^n\sin kx}=\lim_{x \to 0} \frac{(n+1)x^2+\sum_{k=1}^n \frac{k^2x^2}{2}+o(x^2)}{x^2\sum_{k=1}^n k +o(x^2)}=\frac{(n+1)+\frac{n(n+1)(2n+1}{12}}{\frac{n(n+1}{2}}= \frac{2n^2+n+12}{6n}.$

Using $\boxed{\lim_{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta ^2}}=\frac{1}{2},$ we have ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}=\frac{1}{12}n(n+1)(n+2),$ without using Landau sign $o.$ Thus, by $\lim_{x\rightarrow 0} \frac{(1+x^2)^{n+1}-1}{x^2}=\lim_{t\rightarrow 0} \frac{(1+t)^{n+1}-1}{t}$ $=\lim_{t\rightarrow 0}(1+t+t^2+\cdots +t^{n})=n+1,$ $\lim_{x\rightarrow 0} \sum_{k=1}^n \frac{\sin kx}{kx}\cdot k=1+2+\cdots+n=\frac{n(n+1)}{2}.$ Thus $\lim_{x\to 0}\frac{ \displaystyle (1+x^2)^{n+1}-\prod_{k=1}^n\cos kx}{ \displaystyle x\sum_{k=1}^n\sin kx} =\lim_{x\rightarrow 0} \frac{\frac{(1+x^2)^{n+1}-1}{x^2}+\frac{\displaystyle 1-\prod_{k=1}^n\cos kx}{x^2}}{\displaystyle \sum_{k=1}^n \frac{\sin kx}{kx}\cdot k}$ $=\frac{n+1+\frac{n(n+1)(n+2)}{12}}{\frac{n(n+1)}{2}}=\frac{2n^2+n+12}{6n}.\ Q.E.D.$