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JH

# Let $a_n = \left ( 1 + \frac {1}{n} \right)^n.$(a) Show that if $0 \le a < b,$ then $\frac {b^{n + 1} - a^{n + 1}}{b -a } < (n + 1)b^n$(b) Deduce that $b^n[(n + 1)a - nb] < a^{n + 1}.$(c) Use $a = 1 + 1/(n + 1)$ and $b = 1 + 1/n$ in part (b) to show that $\left \{ a_n \right \}$ is increasing.(d) Use $a = 1$ and $b = 1 + 1/(2n)$ in part (b) to show that $a_{2n} < 4.$(e) Use parts (c) and (d) to show that $a_n < 4$ for all $n.$(f) Use Theorem 12 to show that $\lim_{n \to\infty} (1 + 1/n)^n$ exists.(The limit is $e.$ See Equation 3.6.6.)

## (a) $$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1) b^{n}$$(b) $$b^{n}[(n+1) a-n b]<a^{n+1}$$(c) $$a_{n}<a_{n+1}$$(d) $$\left(1+\frac{1}{2 n}\right)^{2 n}<2^{2}=4$$(e) the whole sequence $\left\{a_{n}\right\}$ is bounded by 4(f) Theorem 11 requires that we have a monotonic sequence that is bounded.By part $\mathrm{c},$ we know that $a_{n}$ is monotonic. By part e, we know that $a_{n}$ isbounded. Therefore, the limit exists.

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