Đáp án đúng: C Ta có $0\le \left| \frac{{{\left( -1 \right)}^{n}}}{n+1} \right|\le \frac{1}{n+1}\le \frac{1}{n}\to 0\xrightarrow[{}]{}\lim \frac{{{\left( -1 \right)}^{n}}}{n+1}=0\xrightarrow[{}]{}\lim \left( 4+\frac{{{\left( -1 \right)}^{n}}}{n+1} \right)=4.$ Chọn đáp án C.