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.