Finding limit of $m(1-\frac{1}{m+1})^{m^2-m-1}$ as $m\rightarrow \infty$

Question

Intuitively I feel like the limit is going to 0. But while proving if I take log then it is coming infinity. Please solve the problem.

Answer 1

By root test we have

$$\sqrt[m]{a_m}=\sqrt[m]m\left(1-\frac{1}{m+1}\right)^{\frac{m^2-m-1}m}\to 1\cdot \frac1e=\frac1e<1$$

indeed

$$\sqrt[m]m=e^{\frac{\log m}m}\to e^0=1$$

and

$$\left(1-\frac{1}{m+1}\right)^{\frac{m^2-m-1}m}=\left[\left(1-\frac{1}{m+1}\right)^{m+1}\right]^\frac{m^2-m-1}{m(m+1)}\to \frac1e$$

therefore

$$a_m=m\left(1-\frac{1}{m+1}\right)^{m^2-m-1}\to 0$$