Show that $\frac{\log(n+1)}{n}\le\frac{\log(n)}{n-1}$ for $n\ge2$:
$$
\begin{align}
\left(1+\frac1n\right)^n&\le n+1\tag{1}\\[6pt]
(n+1)^{n-1}&\le n^n\tag{2}\\[12pt]
(n-1)\log(n+1)&\le n\log(n)\tag{3}\\[6pt]
\frac{\log(n+1)}{n}&\le\frac{\log(n)}{n-1}\tag{4}
\end{align}
$$
Explanation:
$\begin{array}{l}
(1):&\text{the left side is $\lt e$}\\
(2):&\text{multiply by $n^n/(n+1)$}\\
(3):&\text{take logs}\\
(4):&\text{divide by $n(n-1)$}
\end{array}$
$$
\begin{align}
\sum_{n=2}^\infty\frac{\log(n)}{n!}