\lim_{x\to \infty } \frac{-x }{e^{(n+1)x}(n+1)}\searrow 0-0+\frac{1}{n+1}\int _0^{\infty } e^{-(n+1) x}dx=\lim_{x\to \infty } \frac{-1}{(n+1)^2}e^{-(n+1) x}+\frac{1}{(n+1)^2}=\frac{1}{(n+1)^2}