Factorials

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Remember our facts about factorials.

The Ratio Test is a useful convergence test that is very useful when trying to determine convergence or divergence of series whose summands involve products of polynomials, factorials, and exponential functions.

A fundamental skill when using this test is to be able to simplify ratios of expressions involving these types of terms.

This assignment will review some of the algebraic skills required when using this test.

I understand. I do not understand.

Ratios of exponentials

Ratios of exponential terms also simplify nicely. Suppose $a_k=2^k$ and $b_k=3^{2k}$ . Simplify: $\frac {a_{k+1}}{a_k} = \answer {2}$ $\frac {b_{k+1}}{b_k}=\answer {9}$

Suppose $a_k$ is a sequence whose $k$ th term is given by: $a_k=4^{k^2}$ Note: The $k^2$ is in the exponent. Simplify $\frac {a_{k+1}}{a_k} = \answer {4^{2k+1}}$

Factorials

Given an integer $n$ , the notation “ $n!$ ” is defined as follows: $n! = n(n-1)(n-2)\dots (3)(2)(1)$ For instance, $3!=3\cdot 2\cdot 1 = 6$ . Compute: $\begin{align*} 4! &= \answer{24}\\ 6! &= \answer{720}\\ \frac{6!}{4!} &= \answer{30} \end{align*}$

How did you compute $\frac {6!}{4!}$ in the last problem? You could certainly find $6!$ and $4!$ separately, then divide, but there is a nicer way to do this: $\begin{align*} \frac{6!}{4!} &= \frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}\\ &=\frac{6\cdot5\cdot\not{4}\cdot\not{3}\cdot\not{2}\cdot\not{1}}{\not{4}\cdot\not{3}\cdot\not{2}\cdot\not{1}}\\ &= 6\cdot 5. \end{align*}$ Ratios of factorials are always easiest to compute by canceling like terms! Compute the following, simplify your final answers. $\begin{align*} \frac{5!}{4!} &= \answer{5}\\ \frac{6!}{3!\cdot 3!} &= \answer{20}\\ \frac{k!}{(k+2)!} &=\answer{\frac{1}{(k+2)(k+1)}} \end{align*}$

Often, factorials arise in sequences, and it is important to understand how to write out various terms. Suppose $a_k$ is a sequence whose $k$ th term is given by: $a_k = (2k+1)!$

(a): $a_2 = \answer {120}$
(b): Find an expression for $a_{k+1}$ . Express your answer in the form $(ck+d)!$ . $a_{k+1} = \left (\answer {2} \cdot k + \answer {3}\right )!$
(c): Calculate and simplify: $\frac {a_{k+1}}{a_k} = \answer {(2k+3)(2k+2)}$

Let $a_k$ be a sequence whose $k$ th term is given by: $a_k = \frac {(2k)!}{(k!)^2}$ Calculate and simplify: $\begin{align*} \frac{a_{k+1}}{a_k} &= \answer{\frac{4k+2}{k+1}}\\ \lim_{k\to\infty} \frac{a_{k+1}}{a_k} &= \answer{4} \end{align*}$