Show that each of the following sequences is monotonic by determining whether it is increasing, nondecreasing, decreasing, or nonincreasing:

\(\{ s_{n}\} =\left\{ {\dfrac{n}{n+1}}\right\} \)

\(\{ s_{n}\} =\left\{ {\dfrac{e^{n}}{n!}}\right\}\)

\(\{ s_{n}\} =\{\ln n\}\)

Solution (a) We use the algebraic difference test. \[ \begin{eqnarray*} s_{n+1}-s_{n} &=& \frac{n+1}{n+2}-\frac{n}{n+1}=\frac{n^{2}+2n+1-n^{2}-2n}{(n+2)(n+1)}\\[5pt] &=&\frac{1}{(n+2)(n+1)}>0 \qquad \hbox{ for all }n\geq 1 \end{eqnarray*} \]

So, \(\{ s_{n}\} \) is an increasing sequence.

(b) When the sequence contains a factorial, the algebraic ratio test is usually easiest to use. \[ \frac{s_{n+1}}{s_{n}}=\frac{\dfrac{e^{n+1}}{(n+1)!}}{\dfrac{e^{n}}{n!}}=\left( {\frac{e^{n+1}}{e^{n}}}\right) \frac{n!}{(n+1)!}=\frac{e}{n+1}<1 \qquad \hbox{ for all }n\geq 2 \]

After the first term, \(\{ s_{n}\} =\left\{ \dfrac{e^{n}}{n!}\right\} \) is a decreasing sequence.

(c) Here, we use the derivative of the related function \(f(x) =\ln x\) of the sequence \(\{ s_{n}\} =\{\ln n\}.\) Since \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}>0\) for all \(x>0\), it follows that \(f\) is an increasing function and so the sequence \(\{\ln n\}\) is an increasing sequence.