Additional Convergence Tests

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.6 – Absolute Convergence and the Ratio and Root Tests

• Section 11.7 – Strategy for Testing Series

Prerequisite Knowledge

Pre-Algebra

Properties of Exponential Expressions

• ${\displaystyle (ab{)}^c=a^c{b}^c}$$\displaystyle (ab)^c=a^cb^c$

• ${\displaystyle {\left(\frac{a}{b}\right)}^c=\frac{a^c}{b^c}}$$\displaystyle \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

• ${\displaystyle {\left(a^b\right)}^c=a^{bc}}$$\displaystyle \left(a^b\right)^c=a^{bc}$

• ${\displaystyle a^b{a}^c=a^{b+c}}$$\displaystyle a^ba^c=a^{b+c}$

• ${\displaystyle \frac{a^b}{a^c}=a^{b-c}}$$\displaystyle \frac{a^b}{a^c}=a^{b-c}$

Calculus I

Limits

Infinite Limits

• If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$, i.e., $f(x)$$f(x)$ becomes large without bound, then the limit does not exist.

• If ${\displaystyle \underset{x\to a}{lim}f(x)=-\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=-\infty$, i.e., $f(x)$$f(x)$ becomes small without bound, then the limit does not exist.

NOTE: $\mathrm{\infty }$$\infty$ is NOT a real number.

Limits at Infinity

Let $f$$f$ be a function defined on some open interval $(a,\mathrm{\infty })$$(a,\infty)$. Then $\underset{x\to \mathrm{\infty }}{lim}f(x)=L$$\lim_{x\rightarrow\infty}{f(x)}=L$ means that as $x$$x$ becomes large without bound, the values of $f(x)$$f(x)$ become arbitrarily close to $L$$L$.

NOTE: Since $\mathrm{\infty }$$\infty$ is NOT a real number, then $f(\mathrm{\infty })$$f(\infty)$ is NOT meaningful.

Indeterminate Forms of Limits

Indeterminate Quotients

1. Type ${\displaystyle \frac{0}{0}}$$\displaystyle\frac{0}{0}$: If ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$, then ${\displaystyle \underset{x\to a}{lim}\frac{f(x)}{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}$ may or may not exist.

2. Type ${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$$\displaystyle\frac{\infty}{\infty}$: If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}\frac{f(x)}{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}$ may or may not exist.

Indeterminate Differences

3. Type ${\displaystyle \mathrm{\infty }-\mathrm{\infty }}$$\displaystyle\infty -\infty$: If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}[f(x)-g(x)]}$$\displaystyle\lim_{x\rightarrow a}{\left[f(x)-g(x)\right]}$ may or may not exist.

Indeterminate Products

4. Type ${\displaystyle 0\cdot \mathrm{\infty }}$$\displaystyle 0\cdot\infty$: If ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}[f(x)g(x)]}$$\displaystyle\lim_{x\rightarrow a}{\left[f(x)g(x)\right]}$ may or may not exist.

Indeterminate Powers

5. Type ${\displaystyle 0^0}$$\displaystyle 0^0$: If ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$, then ${\displaystyle \underset{x\to a}{lim}f(x{)}^{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{f(x)^{g(x)}}$ may or may not exist.

6. Type ${\displaystyle {\mathrm{\infty }}^0}$$\displaystyle \infty^0$: If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$, then ${\displaystyle \underset{x\to a}{lim}f(x{)}^{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{f(x)^{g(x)}}$ may or may not exist.

7. Type ${\displaystyle 1^{\mathrm{\infty }}}$$\displaystyle 1^{\infty}$: If ${\displaystyle \underset{x\to a}{lim}f(x)=1}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=1$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}f(x{)}^{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{f(x)^{g(x)}}$ may or may not exist.

l'Hôpital's Rule

NOTE: l'Hôpital's Rule is valid only for indeterminate quotients.

Definition

Suppose $f$$f$ and $g$$g$ are differentiable and $g^{\prime }(x)\ne 0$$g^{\,\prime}(x)\ne 0$ on some open interval that contains $a$$a$. Suppose that

• ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$; or

• ${\displaystyle \underset{x\to a}{lim}f(x)=\pm \mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\pm\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\pm \mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\pm\infty$

then $\underset{x\to a}{lim}{\displaystyle \frac{f(x)}{g(x)}}=\underset{x\to a}{lim}{\displaystyle \frac{f^{\prime }(x)}{g^{\prime }(x)}}$$\lim_{x\rightarrow a}{\dfrac{f(x)}{g(x)}}=\lim_{x\rightarrow a}{\dfrac{f^{\,\prime}(x)}{g^{\,\prime}(x)}}$

Factorials

For all positive integers, $n$$n$, $n!$$n!$ (read as $n$$n$ factorial) is defined as

$n!=n(n-1)(n-2)\cdots (2)(1)$$n! = n(n-1)(n-2) \cdots (2)(1)$

In general,

$n!=n(n-1)!=n(n-1)(n-2)!=n(n-1)(n-2)(n-3)!=\cdots$$n! = n(n-1)! = n(n-1)(n-2)! = n(n-1)(n-2)(n-3)! = \cdots$

Gamma Function

The factorial is formally defined using the gamma function:

$\mathrm{\Gamma }(n)={\int }_0^{\mathrm{\infty }}{x}^{n-1}{e}^{-x}dx=(n-1)!$$\Gamma(n) = \int_0^{\infty}{x^{n-1}e^{-x}\,dx} = (n-1)!$

Check Your Work

Use Technology to Evaluate Gamma Funciton

Mathematica

​x
1
(* Compute 5! *)
2
5!
3

4
(* Alternatively, compute 5! using the gamma function *)
5
Gamma[6]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Simplifying Rational Expressions with Factorials

Example 01: Simplify: ${\displaystyle \frac{(n-1)!}{n!}}$$\displaystyle \frac{(n-1)!}{n!}$

Solution:

${\displaystyle \frac{(n-1)!}{n!}=\frac{\cancel{(n-1)!}}{n\cancel{(n-1)!}}=\frac{1}{n}}$$\displaystyle \frac{(n-1)!}{n!} = \frac{\cancel{(n-1)!}}{n\cancel{(n-1)!}} = \frac{1}{n}$

Example 02: Simplify: ${\displaystyle \frac{n!}{(n-2)!}}$$\displaystyle \frac{n!}{(n-2)!}$

Solution:

${\displaystyle \frac{n!}{(n-2)!}=\frac{n(n-1)\cancel{(n-2)!}}{\cancel{(n-2)!}}=n(n-1)}$$\displaystyle \frac{n!}{(n-2)!} = \frac{n(n-1)\cancel{(n-2)!}}{\cancel{(n-2)!}} = n(n-1)$

Example 03: Simplify: ${\displaystyle \frac{(n+3)!}{(n+1)!}}$$\displaystyle \frac{(n+3)!}{(n+1)!}$

Solution:

${\displaystyle \frac{(n+3)!}{(n+1)!}=\frac{(n+3)(n+2)\cancel{(n+1)!}}{\cancel{(n+1)!}}=(n+3)(n+2)}$$\displaystyle \frac{(n+3)!}{(n+1)!} = \frac{(n+3)(n+2)\cancel{(n+1)!}}{\cancel{(n+1)!}} = (n+3)(n+2)$

Example 04: Simplify: ${\displaystyle \frac{n(n-2)!}{(n+1)!}}$$\displaystyle \frac{n(n-2)!}{(n+1)!}$

Solution:

${\displaystyle \frac{n(n-2)!}{(n+1)!}=\frac{\bcancel{n}\cancel{(n-2)!}}{(n+1)\bcancel{n}(n-1)\cancel{(n-2)!}}=\frac{1}{(n+1)(n-1)}}$$\displaystyle \frac{n(n-2)!}{(n+1)!} = \frac{\bcancel{n}\cancel{(n-2)!}}{(n+1)\bcancel{n}(n-1)\cancel{(n-2)!}} = \frac{1}{(n+1)(n-1)}$

Example 05: Simplify: ${\displaystyle \frac{(2n)!}{(2n+2)!}}$$\displaystyle \frac{(2n)!}{(2n+2)!}$

Solution:

${\displaystyle \frac{(2n)!}{(2n+2)!}=\frac{\cancel{(2n)!}}{(2n+2)(2n+1)\cancel{(2n)!}}=\frac{1}{(2n+2)(2n+1)}}$$\displaystyle \frac{(2n)!}{(2n+2)!} = \frac{\cancel{(2n)!}}{(2n+2)(2n+1)\cancel{(2n)!}} = \frac{1}{(2n+2)(2n+1)}$

Example 06: Simplify: ${\displaystyle \frac{(2n)!}{(2[n-1])!}}$$\displaystyle \frac{(2n)!}{(2[n-1])!}$

Solution:

${\displaystyle \frac{(2n)!}{(2[n-1])!}=\frac{(2n)!}{(2n-2)!}=\frac{2n(2n-1)\cancel{(2n-2)!}}{\cancel{(2n-2)!}}=2n(2n-1)}$$\displaystyle \frac{(2n)!}{(2[n-1])!} = \frac{(2n)!}{(2n-2)!} = \frac{2n(2n-1)\cancel{(2n-2)!}}{\cancel{(2n-2)!}} = 2n(2n-1)$

Check Your Work

Use Technology to Simplify Factorials

Mathematica

xxxxxxxxxx
2
1
(* Example 06 *)
2
FullSimplify[(2n)!/(2(n-1))!]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Last Modified: Tuesday, 17 November 2020 6:42 EDT