B. (Modification of Exercise 3.4.2) Prove that the following sequence is decreasing and bounded below: $$\left\langle \frac{2^{n+3}}{(n+3)!} \right\rangle$$
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Let $a_n = \frac{2^{n+3}}{(n+3)!}$. Then $a_{n+1} = \frac{2^{(n+1)+3}}{((n+1)+3)!} = \frac{2^{n+4}}{(n+4)!}$. We want to show that $\frac{a_{n+1}}{a_n} \le 1$. $$\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+4}}{(n+4)!}}{\frac{2^{n+3}}{(n+3)!}} = Show more…
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