Determine if the sequence $\{ s_{n}\} =\left\{ {\dfrac{2^{n}}{n!}}\right\}$ converges or diverges.

Solution To see if $\left\{ \dfrac{2^{n}}{n!}\right\}$ is monotonic, find the algebraic ratio $\dfrac{s_{n+1}}{s_{n}}$: $$\frac{s_{n+1}}{s_{n}}=\frac{\dfrac{2^{n+1}}{(n+1)!}}{\dfrac{2^{n}}{n!}}=\frac{2^{n+1}\,n!}{(n+1)!\,2^{n}}=\frac{2}{n+1}\leq 1 \qquad \hbox{ for all }n \geq 1$$

Since $s_{n+1}\leq s_{n}$ for $n\geq 1,$ the sequence $\{ s_{n}\}$ is nonincreasing.

Next, since each term of the sequence is positive, $s_{n}>0$ for $n\geq 1,$ the sequence $\{ s_{n}\}$ is bounded from below.

Since $\{ s_{n}\}$ is nonincreasing and bounded from below, it converges.