Equivalence between $\sum u_n$ and $\sum\ln(1+u_n)$

1. Equivalence of the two series. Both series have nonnegative terms, so we can use the equivalence test.

• Suppose $\sum u_n$ converges. Then $u_n\to0$, hence $\ln(1+u_n)\sim u_n$. Since the terms are nonnegative and equivalent, $\sum\ln(1+u_n)$ converges.

• Conversely, suppose $\sum\ln(1+u_n)$ converges. Then $\ln(1+u_n)\to0$, hence $u_n\to0$ (as $u_n=e^{\ln(1+u_n)}-1\to0$). Therefore $\ln(1+u_n)\sim u_n$, and by the equivalence test $\sum u_n$ converges.

2. Convergence of $\sum v_n$. By 1., it suffices to study the series with nonnegative term $u_n=e^{-n}\dfrac{n!}{n^n}$. Compute the ratio:
$$\frac{u_{n+1}}{u_n}=e^{-(n+1)}\frac{(n+1)!}{(n+1)^{n+1}}\cdot e^{n}\frac{n^n}{n!}=e^{-1}\left(\frac{n}{n+1}\right)^n=e^{-1}\left(1+\frac1n\right)^{-n}.$$
Since $\left(1+\frac1n\right)^n\to e$, we get
$$\frac{u_{n+1}}{u_n}\longrightarrow e^{-1}\cdot e^{-1}=e^{-2}<1.$$
By d'Alembert's ratio test, $\sum u_n$ converges. By 1., $\sum v_n$ converges.