Answer
Converges Absolutely
Work Step by Step
Here, $a_n=\dfrac{n+1}{n!}$ and $a_{n+1}=\dfrac{(n+1)+1}{(n+1)!}$
Apply Ratio Test.
$\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} |\dfrac{\dfrac{(n+2)}{(n+1)!}}{\dfrac{n+1}{n!}}|$
Now, $\lim\limits_{n \to \infty} |\dfrac{\dfrac{n+2}{n!(n+1)!}}{\dfrac{n+1}{n!}}|=\lim\limits_{n \to \infty} \dfrac{(n+2)}{(n+1)^2}$
Thus, $\lim\limits_{n \to \infty} \dfrac{1}{(n+1)}+\lim\limits_{n \to \infty} \dfrac{1}{(n+1)^2}=\dfrac{1}{\infty}+\dfrac{1}{\infty}=0 \lt 1$
Hence, the series Converges Absolutely by the Ratio Test.
