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Calculus 2 : Series in Calculus

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 1}^{\infty} \frac{11(n-1)!}{ln(n))}$ $\displaystyle \sum_{n = 1}^{\infty} \frac{11(n-1)!}{ln(n))}$

Note: $f(n) = \frac{(n-1)!}{ln(n)}$ $\displaystyle f(n) = \frac{(n-1)!}{ln(n)}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{11}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{11}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 11$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 11$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$ $\displaystyle Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Explanation:

Determine the convergence of the series based on the limits.

$\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ $\displaystyle \lim_{n\rightarrow \infty}$ .

$\frac{\frac{(n)!}{ln(n+1)}} {\frac{(n-1)!}{ln(n)}}$ $\displaystyle \frac{\frac{(n)!}{ln(n+1)}} {\frac{(n-1)!}{ln(n)}}$ $= \frac{\frac{(n)!}{(n-1)!}}{\frac{ln(n+1)}{ln(n)}}$ $\displaystyle = \frac{\frac{(n)!}{(n-1)!}}{\frac{ln(n+1)}{ln(n)}}$ $= \frac{nln(n)}^{ln(n+1)}$ $\displaystyle = \frac{nln(n)}^{ln(n+1)}$ $,\lim_{n\rightarrow \infty} \frac{nln(n)}^{ln(n+1)}$ $\rightarrow \infty$ $\displaystyle \rightarrow \infty$

Note: $\displaystyle nln(n) > ln(n+1)$

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Example Question #61 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 1}^{\infty} \frac{11^n}{ln(n)}$ $\displaystyle \sum_{n = 1}^{\infty} \frac{11^n}{ln(n)}$

Note: $f(n) = \frac{11^n}{ln(n)}$ $\displaystyle f(n) = \frac{11^n}{ln(n)}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 11$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 11$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$ $\displaystyle Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 11$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 11$

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ $\displaystyle \lim_{n\rightarrow \infty}$ .

$\frac{\frac{11^{n+1}}{ln(n+1)}} {\frac{11^n}{ln(n)}}$ $\displaystyle \frac{\frac{11^{n+1}}{ln(n+1)}} {\frac{11^n}{ln(n)}}$ $= \frac{\frac{11^{n+1}}{11^n}}{\frac{ln(n+1)}{ln(n)}}$ $\displaystyle = \frac{\frac{11^{n+1}}{11^n}}{\frac{ln(n+1)}{ln(n)}}$

$= \frac{11ln(n)}^{ln(n+1)}$ $,\lim_{n\rightarrow \infty}$ $\displaystyle ,\lim_{n\rightarrow \infty}$ $\frac{11ln(n)}^{ln(n+1)}$ $\displaystyle \frac{11ln(n)}^{ln(n+1)}$ = 11 $\displaystyle = 11$

Note: $\lim_{n\rightarrow \infty} \frac{ln(n)}{ln(n+1)}=1$ $\displaystyle \lim_{n\rightarrow \infty} \frac{ln(n)}{ln(n+1)}=1$

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Example Question #26 : Ratio Test

For the following series

$\small \sum_{n=0}^{\infty}\frac{n!x^n}{(2n)!}$ $\displaystyle \small \sum_{n=0}^{\infty}\frac{n!x^n}{(2n)!}$

what is its radius of convergence?

Possible Answers:

$\small 0$ $\displaystyle \small 0$

$\small \frac{1}{\pi}$ $\displaystyle \small \frac{1}{\pi}$

$\small \infty$ $\displaystyle \small \infty$

$\small 3!$ $\displaystyle \small 3!$

$\small 1$ $\displaystyle \small 1$

Correct answer:

$\small \infty$ $\displaystyle \small \infty$

Explanation:

We can use the ratio test to find the radius of convergence:

$\small \small \small \small \small\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right |= \lim \left| \frac{x^{n+1}(n+1)!}{(2(n+1))!} \frac{(2n)!}{n!x^n} \right|=\lim \left| \frac{x^{n+1}(n+1)!}{(2n+2)!} \frac{(2n)!}{n!x^n} \right|$ $\displaystyle \small \small \small \small \small\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right |= \lim \left| \frac{x^{n+1}(n+1)!}{(2(n+1))!} \frac{(2n)!}{n!x^n} \right|=\lim \left| \frac{x^{n+1}(n+1)!}{(2n+2)!} \frac{(2n)!}{n!x^n} \right|$

$\small \small =\lim|x| \left| \frac{(n+1)!}{n!} \frac{(2n)!}{(2n+2)!} \right|=\lim|x| \left| \frac{n+1}{(2n+1)(2n+2)} \right|=0<1$ $\displaystyle \small \small =\lim|x| \left| \frac{(n+1)!}{n!} \frac{(2n)!}{(2n+2)!} \right|=\lim|x| \left| \frac{n+1}{(2n+1)(2n+2)} \right|=0< 1$

where the limit is independent of $\small x$ $\displaystyle \small x$ . This means the series converges for all $\small x\in\mathbb{R}$ $\displaystyle \small x\in\mathbb{R}$ , so the radius of convergence is $\small \infty$ $\displaystyle \small \infty$ .

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Example Question #1 : Sequences

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$ $\displaystyle \small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

Possible Answers:

$\small \small R=\frac{1}{4}$ $\displaystyle \small \small R=\frac{1}{4}$

$\small \small R=2$ $\displaystyle \small \small R=2$

$\small \small R=\infty$ $\displaystyle \small \small R=\infty$

$\small R=4$ $\displaystyle \small R=4$

$\small R=0$ $\displaystyle \small R=0$

Correct answer:

$\small R=4$ $\displaystyle \small R=4$

Explanation:

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$ $\displaystyle \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$ $\displaystyle \small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$ $\displaystyle \small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}< 1$

$\small \implies |x|<4=R$ $\displaystyle \small \implies |x|< 4=R$

This means the radius of convergence is $\small R=4$ $\displaystyle \small R=4$ .

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Example Question #2842 : Calculus Ii

Find the radius of convergence of the following power series.

$\small \sum_{n=0}^{\infty} \frac{(n!)^3x^{4n}}{(3n)!}$ $\displaystyle \small \sum_{n=0}^{\infty} \frac{(n!)^3x^{4n}}{(3n)!}$

Possible Answers:

$\small \small R=0$ $\displaystyle \small \small R=0$

$\small R=27^{1/4}$ $\displaystyle \small R=27^{1/4}$

$\small \small R=9^{1/4}$ $\displaystyle \small \small R=9^{1/4}$

$\small \small R=\frac{1}{27^{1/4}}$ $\displaystyle \small \small R=\frac{1}{27^{1/4}}$

$\small R=\infty$ $\displaystyle \small R=\infty$

Correct answer:

$\small R=27^{1/4}$ $\displaystyle \small R=27^{1/4}$

Explanation:

To find the radius of convergence of

$\small \sum_{n=0}^{\infty} \frac{(n!)^3x^{4n}}{(3n)!}$ $\displaystyle \small \sum_{n=0}^{\infty} \frac{(n!)^3x^{4n}}{(3n)!}$

we can use the limit from the ratio test:

$\small \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim\left| \frac{x^{4n+4}[(n+1)!]^3}{(3n+3)!}\frac{(3n)!}{x^{4n}(n!)^3} \right |$ $\displaystyle \small \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim\left| \frac{x^{4n+4}[(n+1)!]^3}{(3n+3)!}\frac{(3n)!}{x^{4n}(n!)^3} \right |$

$\small \small \small =\lim |x^4|\left| \frac{[(n+1)!]^3}{(n!)^3}\frac{(3n)!}{(3n+3)!} \right |=\lim |x^4| \left|\frac{(n+1)(n+1)(n+1)}{(3n+1)(3n+2)(3n+3)} \right |$ $\displaystyle \small \small \small =\lim |x^4|\left| \frac{[(n+1)!]^3}{(n!)^3}\frac{(3n)!}{(3n+3)!} \right |=\lim |x^4| \left|\frac{(n+1)(n+1)(n+1)}{(3n+1)(3n+2)(3n+3)} \right |$

$\small =\frac{|x^4|}{27}<1$ $\displaystyle \small =\frac{|x^4|}{27}< 1$

$\small \small \iff |x|<27^{1/4}=R$ $\displaystyle \small \small \iff |x|< 27^{1/4}=R$

So the radius of convergence of the power series mentioned is $\small R=27^{1/4}$ $\displaystyle \small R=27^{1/4}$ .

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Example Question #61 : Series In Calculus

Find the radius of convergence of the power series:

$\small \small \sum_{n=0}^\infty \frac{n! (2n+1)!}{(3n)!}x^{2n}$ $\displaystyle \small \small \sum_{n=0}^\infty \frac{n! (2n+1)!}{(3n)!}x^{2n}$

Possible Answers:

$\small \small R= \infty$ $\displaystyle \small \small R= \infty$

$\small R= \frac{\sqrt{27}}{2}$ $\displaystyle \small R= \frac{\sqrt{27}}{2}$

$\small \frac{27}{4}$ $\displaystyle \small \frac{27}{4}$

$\small \small R= 0$ $\displaystyle \small \small R= 0$

$\small \small R= \sqrt{\frac{4}{27}}$ $\displaystyle \small \small R= \sqrt{\frac{4}{27}}$

Correct answer:

$\small R= \frac{\sqrt{27}}{2}$ $\displaystyle \small R= \frac{\sqrt{27}}{2}$

Explanation:

We can use the ratio test limit to find the radius of convergence:

$\small \small \lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n} \right |=\lim\left|\frac{(n+1)! (2n+3)!x^{2n+2}}{(3n+3)!}\frac{(3n)!}{(2n+1)!n!x^{2n}} \right |$ $\displaystyle \small \small \lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n} \right |=\lim\left|\frac{(n+1)! (2n+3)!x^{2n+2}}{(3n+3)!}\frac{(3n)!}{(2n+1)!n!x^{2n}} \right |$

$\small =\lim|x^2|\left|\frac{(n+1)!}{n!}\frac{(2n+3)!}{(2n+1)!}\frac{(3n)!}{(3n+3)!} \right |=\lim|x^2|\left|\frac{(n+1)(2n+2)(2n+3)}{(3n+1)(3n+2)(3n+3)} \right |$ $\displaystyle \small =\lim|x^2|\left|\frac{(n+1)!}{n!}\frac{(2n+3)!}{(2n+1)!}\frac{(3n)!}{(3n+3)!} \right |=\lim|x^2|\left|\frac{(n+1)(2n+2)(2n+3)}{(3n+1)(3n+2)(3n+3)} \right |$

$\small =\frac{4|x^2|}{27}<1$ $\displaystyle \small =\frac{4|x^2|}{27}< 1$

$\small \small \iff |x| < \sqrt{\frac{27}{4}}=\frac{\sqrt{27}}{2}=R$ $\displaystyle \small \small \iff |x| < \sqrt{\frac{27}{4}}=\frac{\sqrt{27}}{2}=R$

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Example Question #65 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{4^n}{10^{3n}}\newline note: \hspace f(n) = \frac{4^n}{10^{3n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1000$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1000$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$ $\displaystyle Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{1000}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{1000}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{1000}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{1000}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{4^{\left(n+1\right)}}{10^{3\left(n+1\right)}}}{\frac{4^n}{10^{3n}}} = \frac{4^{n+1}\cdot \:10^{3n}}{4^n\cdot \:10^{3\left(n+1\right)}}= \frac{4}{1000}$ $\displaystyle \frac{\frac{4^{\left(n+1\right)}}{10^{3\left(n+1\right)}}}{\frac{4^n}{10^{3n}}} = \frac{4^{n+1}\cdot \:10^{3n}}{4^n\cdot \:10^{3\left(n+1\right)}}= \frac{4}{1000}$

$\lim_{n\rightarrow \infty} \frac{4}{1000} = \frac{4}{1000}$ $\displaystyle \lim_{n\rightarrow \infty} \frac{4}{1000} = \frac{4}{1000}$

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Example Question #61 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11^n}{6^{3n}}\newline note: \hspace f(n) = \frac{11^n}{6^{3n}}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$ $\displaystyle Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{3}$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{3}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{216}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{216}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{216}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{216}$

Explanation:

$\frac{\frac{11^{\left(n+1\right)}}{6^{3\left(n+1\right)}}}{\frac{11^n}{6^{3n}}} = \frac{6^{3n}\cdot \:11^{n+1}}{6^{3\left(n+1\right)}\cdot \:11^n} = \frac{11}{216}$ $\displaystyle \frac{\frac{11^{\left(n+1\right)}}{6^{3\left(n+1\right)}}}{\frac{11^n}{6^{3n}}} = \frac{6^{3n}\cdot \:11^{n+1}}{6^{3\left(n+1\right)}\cdot \:11^n} = \frac{11}{216}$

$\lim_{n\rightarrow \infty} \frac{11}{216} = \frac{11}{216}$ $\displaystyle \lim_{n\rightarrow \infty} \frac{11}{216} = \frac{11}{216}$

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Example Question #31 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{10\cdot \:7^n}{2^{3n}}\newline note: \hspace f(n) = \frac{10\cdot \:7^n}{2^{3n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{8}{7}$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{8}{7}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{8}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{8}$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$ $\displaystyle Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{8}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{8}$

Explanation:

$\frac{\frac{7^{n+1}}{2^{3\left(n+1\right)}}}{\frac{7^n}{2^{3n}}} = \frac{2^{3n}\cdot \:7^{n+1}}{2^{3\left(n+1\right)}\cdot \:7^n} = \frac{7}{8}$ $\displaystyle \frac{\frac{7^{n+1}}{2^{3\left(n+1\right)}}}{\frac{7^n}{2^{3n}}} = \frac{2^{3n}\cdot \:7^{n+1}}{2^{3\left(n+1\right)}\cdot \:7^n} = \frac{7}{8}$

$\lim_{n\rightarrow \infty}\frac{7}{8} =\frac{7}{8}$ $\displaystyle \lim_{n\rightarrow \infty}\frac{7}{8} =\frac{7}{8}$

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Example Question #32 : Ratio Test

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11\cdot \:20^n}{4^{2n}}\newline note: \hspace f(n) = \frac{11\cdot \:20^n}{4^{2n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$ $\displaystyle Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{4}$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{4}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{20}$ $\displaystyle Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{11}{20}$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{4}$ $\displaystyle Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{4}$

Explanation:

$\frac{\frac{20^{\left(n+1\right)}}{4^{2\left(n+1\right)}}}{\frac{20^n}{4^{2n}}} = \frac{4^{2n}\cdot \:20^{n+1}}{4^{2\left(n+1\right)}\cdot \:20^n} = \frac{20}{16} = \frac{5}{4}$ $\displaystyle \frac{\frac{20^{\left(n+1\right)}}{4^{2\left(n+1\right)}}}{\frac{20^n}{4^{2n}}} = \frac{4^{2n}\cdot \:20^{n+1}}{4^{2\left(n+1\right)}\cdot \:20^n} = \frac{20}{16} = \frac{5}{4}$

$\lim_{n\rightarrow \infty} \frac{5}{4} = \frac{5}{4}$ $\displaystyle \lim_{n\rightarrow \infty} \frac{5}{4} = \frac{5}{4}$

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Calculus 2 : Series in Calculus

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