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Ratio test Quiz 2

mathematics

 Test the convergence of the series $\sum{\dfrac{{{3}^{n}}-1}{{{5}^{n}}+1}}$ using ratio test.

 To test the convergence of a series using a ratio test, first find the $n^{th}$ and $(n+1)^{th}$ term. Then find the limit, $L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \text{If L  <  1, the series converges} \\ \text{If L  >  1, the series diverges} \\ \text{If L = 1, the test fails}$

  $\text{For a ratio test, first find the }n^{th}\text{ and }(n+1)^{th}\text{ term.} \\ {{a}_{n}}=\dfrac{{{3}^{n}}-1}{{{5}^{n}}+1} ,{{a}_{n+1}}=\dfrac{{{3}^{n+1}}-1}{{{5}^{n+1}}+1} \\ \text{To test the convergence, }L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\ \therefore L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{\dfrac{{{3}^{n+1}}-1}{{{5}^{n+1}}+1}}{\dfrac{{{3}^{n}}-1}{{{5}^{n}}+1}} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{3}^{n+1}}-1}{{{5}^{n+1}}+1}\dfrac{{{5}^{n}}+1}{{{3}^{n}}-1} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{3}^{n+1}}{{5}^{n}}\left( 1-\dfrac{1}{{{3}^{n+1}}} \right)\left( 1+\dfrac{1}{{{5}^{n}}} \right)}{{{5}^{n+1}}{{3}^{n}}\left( 1+\dfrac{1}{{{5}^{n+1}}} \right)\left( 1-\dfrac{1}{{{3}^{n}}} \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{3\left( 1-\dfrac{1}{{{3}^{n+1}}} \right)\left( 1+\dfrac{1}{{{5}^{n}}} \right)}{5\left( 1+\dfrac{1}{{{5}^{n+1}}} \right)\left( 1-\dfrac{1}{{{3}^{n}}} \right)} \right| \\ \Rightarrow \dfrac{3}{5} < 1 \\ \therefore \text{ The series }\sum{\dfrac{{{3}^{n}}-1}{{{5}^{n}}+1}}\text{ converges as the limit is equal to }\dfrac{3}{5} < 1$.

 Test the convergence of the series $\dfrac{1.3}{3}+\dfrac{2.4}{5}+\dfrac{3.5}{7}+\dfrac{4.6}{9}+......$ using ratio test.

 To test the convergence of a series using ratio test, first find the $n^{th}$ and $(n+1)^{th}$ term. Then find the limit, $L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|\\ \text{If L  <  1, the series converges} \\ \text{If L  >  1, the series diverges} \\ \text{If L = 1, the test fails}$

  $\text{For a ratio test, first find the }n^{th}\text{ and }(n+1)^{th}\text{ term.} \\ {{a}_{n}}=\dfrac{n\left( n+2 \right)}{2n+1}, {{a}_{n+1}}=\dfrac{\left( n+1 \right)\left( n+3 \right)}{2\left( n+1 \right)+1}=\dfrac{\left( n+1 \right)\left( n+3 \right)}{2n+3} \\ \text{To test the convergence, }L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\ \therefore L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{\left( n+1 \right)\left( n+3 \right)}{2n+3}\times \dfrac{2n+1}{n\left( n+2 \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{n}^{2}}\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{3}{n} \right)\left( 2+\dfrac{1}{n} \right)}{n\left( 2+\dfrac{3}{n} \right)\left( 1+\dfrac{2}{n} \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{n\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{3}{n} \right)\left( 2+\dfrac{1}{n} \right)}{\left( 2+\dfrac{3}{n} \right)\left( 1+\dfrac{2}{n} \right)} \right| \\ \Rightarrow \infty  > 1 \\ \therefore \text{ The series }\dfrac{1.3}{3}+\dfrac{2.4}{5}+\dfrac{3.5}{7}+\dfrac{4.6}{9}+......\text{ diverges as its limit tends to }\infty  > 1$.

 Test the convergence of the series $\sum\limits_{n=1}^{\infty }{\dfrac{{{3}^{3-2n}}}{{{n}^{2}}+5}}$ using ratio test.

 To test the convergence of a series using ratio test, first find the $n^{th}$ and $(n+1)^{th}$ termThen find the limit, $L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|\\ \text{If L  <  1, the series converges} \\ \text{If L  >  1, the series diverges} \\ \text{If L = 1, the test fails}$

  $\text{For a ratio test, first find the }n^{th}\text{ and }(n+1)^{th}\text{ term.} \\ {{a}_{n}}=\dfrac{{{3}^{3-2n}}}{{{n}^{2}}+5}, {{a}_{n+1}}=\dfrac{{{3}^{3-2\left( n+1 \right)}}}{{{\left( n+1 \right)}^{2}}+5} \\ \Rightarrow \dfrac{{{3}^{1-2n}}}{{{\left( n+1 \right)}^{2}}+5} \\ \text{To test the convergence, }L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\ \therefore L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{3}^{1-2n}}}{{{\left( n+1 \right)}^{2}}+5}\dfrac{{{n}^{2}}+5}{{{3}^{3-2n}}} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{3\left( {{3}^{-2n}} \right)}{{{\left( n+1 \right)}^{2}}+5}\dfrac{{{n}^{2}}+5}{{{3}^{3}}\left( {{3}^{-2n}} \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{n}^{2}}+5}{{{3}^{2}}{{\left( n+1 \right)}^{2}}+5} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{1}{9}\dfrac{\left( 1+\dfrac{5}{{{n}^{2}}} \right)}{\left( 1+\dfrac{5}{{{\left( n+1 \right)}^{2}}} \right)} \right| \\ \Rightarrow \dfrac{1}{9} < 1 \\ \therefore \text{ The series }\sum\limits_{n=1}^{\infty }{\dfrac{{{3}^{3-2n}}}{{{n}^{2}}+5}}\text{ converges as the limit is equal to }\dfrac{1}{9} < 1$.

 Test the convergence of the series$\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{2n+1}}}{5n+6}}$ using ratio test.

 To test the convergence of a series using ratio test, first find the $n^{th}$ and $(n+1)^{th}$ term. Then find the limit, $L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\ \text{If L  <  1, the series converges} \\ \text{If L  >  1, the series diverges} \\ \text{If L = 1, the test fails}$

  $\text{For a ratio test, first find the }n^{th}\text{ and }(n+1)^{th}\text{ term.} \\ {{a}_{n}}=\dfrac{{{\left( -1 \right)}^{2n+1}}}{5n+6}, {{a}_{n+1}}=\dfrac{{{\left( -1 \right)}^{2\left( n+1 \right)+1}}}{5\left( n+1 \right)+6}=\dfrac{{{\left( -1 \right)}^{2n+3}}}{5n+11} \\ \text{To test the convergence, }L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\ \therefore L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{\left( -1 \right)}^{2n+3}}}{5n+11}\dfrac{5n+6}{{{\left( -1 \right)}^{2n+1}}} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{\left( -1 \right)}^{2n}}{{\left( -1 \right)}^{3}}}{5n+11}\dfrac{5n+6}{{{\left( -1 \right)}^{2n}}\left( -1 \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{5n+6}{5n+11} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{5+\dfrac{6}{n}}{5+\dfrac{11}{n}} \right|=1 \\ \therefore \text{ The test fails for the series }\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{2n+1}}}{5n+6}}\text{ as the limit is equal to 1.}$

 Test the convergence of the series $\sum\limits_{n=2}^{\infty }{\dfrac{{{\left( -2 \right)}^{3n+1}}\left( n+2 \right)}{{{n}^{3}}{{.5}^{\left( n+1 \right)}}}}$ using ratio test.

  $\text{For a ratio test, first find the }n^{th}\text{ and }(n+1)^{th}\text{ term.} \\ {{a}_{n}}=\dfrac{{{\left( -2 \right)}^{3n+1}}\left( n+2 \right)}{{{n}^{3}}{{.5}^{\left( n+1 \right)}}}, {{a}_{n+1}}=\dfrac{{{\left( -2 \right)}^{3\left( n+1 \right)+1}}\left( \left( n+1 \right)+2 \right)}{{{\left( n+1 \right)}^{3}}{{.5}^{\left( \left( n+1 \right)+1 \right)}}} \\ \Rightarrow \dfrac{{{\left( -2 \right)}^{3n+4}}\left( n+3 \right)}{{{\left( n+1 \right)}^{3}}{{.5}^{\left( n+2 \right)}}} \\ \text{To test the convergence, }L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\ \therefore L=\underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{\left( -2 \right)}^{3n+4}}\left( n+3 \right)}{{{\left( n+1 \right)}^{3}}{{.5}^{\left( n+2 \right)}}}\dfrac{{{n}^{3}}{{.5}^{\left( n+1 \right)}}}{{{\left( -2 \right)}^{3n+1}}\left( n+2 \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{\left( -2 \right)}^{3n}}{{\left( -2 \right)}^{4}}\left( n+3 \right)}{{{\left( n+1 \right)}^{3}}{{.5}^{n}}{{.5}^{2}}}\dfrac{{{n}^{3}}{{.5}^{n}}.5}{{{\left( -2 \right)}^{3n}}\left( -2 \right)\left( n+2 \right)} \right| \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left| \dfrac{-8\left( n+3 \right)}{5{{\left( n+1 \right)}^{3}}}\dfrac{{{n}^{3}}}{\left( n+2 \right)} \right| \\ \Rightarrow \dfrac{8}{5} > 1 \\ \therefore \text{ The series }\sum\limits_{n=2}^{\infty }{\dfrac{{{\left( -2 \right)}^{3n+1}}\left( n+2 \right)}{{{n}^{3}}{{.5}^{\left( n+1 \right)}}}}\text{ diverges as its limit is equal to }\dfrac{8}{5} > 1$