Ratio and Root Test
- Page ID
- 219484
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Ratio Test
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Theorem: The Ratio Test
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Proof: Suppose that
\( \lim\limits_{n\to \infty} |\frac{a_{n+1}}{a_n}| < R < 1 \)
then for the tail,
|an+1| < R |an|
|an+2| < R |an+1| < R2 |an| ...
|an+k| < Rk |an|
So that
\( \displaystyle\sum_{n=1}^{\infty} |a_n| < \displaystyle\sum_{n=1}^{\infty} aR^k \)
Which converges by the GST.
Example
Determine the convergence or divergence of
\( \displaystyle\sum_{n=1}^{\infty} \frac{4^n}{n!} \)
We use the Ratio Test:
\( \lim\limits_{n\to \infty} \frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n} = \lim\limits_{n\to \infty} \frac{4}{n+1} = 0 \)
Hence the series converges by the Ratio Test
Exercises
Determine the convergence or divergence of
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\( \displaystyle\sum_{n=1}^{\infty} \frac{n!}{(n+1)3^n} \)
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\( \displaystyle\sum_{n=1}^{\infty} \frac{2^n}{(n+2)3^{n+1}} \)
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The Root Test
The final test for convergence of a series is called the Root Test.
The Root Test
Let \( \displaystyle\sum_{n=1}^{\infty} a_n \) be a series with nonzero terms at the tail, then
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\( \displaystyle\sum_{n=1}^{\infty} a_n \) converges absolutely if
\[ \lim\limits_{n\to \infty} \sqrt[n]{a_n} < 1 \] -
\( \displaystyle\sum_{n=1}^{\infty} a_n \) diverges if
\[ \lim\limits_{n\to \infty} \sqrt[n]{a_n} > 1 \] -
If
\[ \lim\limits_{n\to \infty} \sqrt[n]{a_n} = 1 \]
then try another test.
Example
Determine the convergence or divergence of\( \displaystyle\sum_{n=1}^{\infty}(\frac{3n}{n+3})^n \)
Hence the series diverges.
We use the root test:
\( \lim\limits_{n\to \infty} \sqrt[n]{(\frac{3n}{n+3})^n} = \lim\limits_{n\to \infty}(\frac{3n}{n+3}) \)
\( \lim\limits_{n\to \infty} \frac{3}{1} = 3 \)
Exercises
Determine the convergence or divergence of
A. \( \displaystyle\sum_{n=1}^{\infty} e^{-n^3-n^{n^2}} \)
B. \( \displaystyle\sum_{n=1}^{\infty} (\frac{n}{n+1})^n\) -