Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is the second of our many quizzes on Direct comparison test. Click on this to access more such competitive quizzes on Algebra.

Direct comparison test Quiz 2

mathematics

 Using Direct Comparison test, Test the convergence or divergence of the series $\sum\limits_{n=1}^{\infty }{\dfrac{{{e}^{-2n}}}{2{{n}^{2}}+4}}$.

 Use the formula $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|}$ if the value is finite then the series is convergent and if the value is infinite or approaches infinity then it is divergent.

  Let’s check for bigger series $ \\ $ $\sum\limits_{n=1}^{\infty }{\dfrac{{{e}^{-2n}}}{2{{n}^{2}}+4}} \\$ If a bigger series converges then smaller converges too. $\\ \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{e}^{-2n-2}}\times {{n}^{2}}}{{{\left( n+1 \right)}^{2}}{{e}^{-2n}}} \\ \underset{n\to \infty }{\mathop{=\lim }}\,\dfrac{{{n}^{2}}}{{{\left( n+1 \right)}^{2}}{{e}^{2}}}=\dfrac{1}{{{e}^{2}}} \\ $ So convergent $\therefore \sum\limits_{n=1}^{\infty }{\dfrac{{{e}^{-2n}}}{2{{n}^{2}}+4}}$ also converges.

 Find which of the following series diverges using a comparison test.

 Check for the values of each and every function and see whether the  value of $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|}$ is finite or not and decides the nature of the series.

  By using alternating series test: $ \\ (a) \sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{2n+3}}}\dfrac{\cos (n)}{{{\left( n+1 \right)}^{2}}}$ $ \\ $ $\underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=0,  {{b}_{n}}$  is decreasing ,finite value so converging. $ \\ (b) \sum\limits_{n=1}^{\infty }{\dfrac{2{{n}^{2}}+4}{\sqrt{n\left( n+1 \right)}}} \\ \underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=0 \\ $ $ {{b}_{n}}=0, {{b}_{n}}$ is decreasing but: $ \\ \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|} \\ =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{2{{\left( n+1 \right)}^{2}}+4}{2{{n}^{2}}+4}\times \dfrac{\sqrt{n}\sqrt{\left( n+1 \right)}}{\sqrt{n+2}\sqrt{\left( n+1 \right)}} \\ =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\sqrt{n}}{\sqrt{\left( n+1 \right)}}\left( 1+\dfrac{2n+1}{{{n}^{2}}+2} \right) \\ \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|}>1 \\$ Since it does not have finite value and hence the function is converging. $\\ (c) \sum\limits_{n=1}^{\infty }{\dfrac{\left( n+1 \right){{e}^{-n}}}{\left( 2n+3 \right)\left( 2n+5 \right)}} \\ {{b}_{n}}=\dfrac{\left( n+1 \right)}{\left( 2n+3 \right)\left( 2n+5 \right)} \\ \underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=0 \\ {{b}_{n}}$ is decreasing and hence has a finite value thus converging. $\\ (d) \sum\limits_{n=1}^{\infty }{\dfrac{\sin (n)}{{{n}^{2}}}{{\left( -1 \right)}^{2n-1}}} \\ \underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=0 \\ {{b}_{n}}$ is decreasing and hence has a finite value thus converging

 Determine the nature of the series $\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{7n+3}}}{2{{n}^{2}}+4n+16}}$

 Check the value of $\underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}$ if the value is finite then it is convergent and if it has an infinite value than it is said to be diverging.

  $\text{Alternating series test: } \\ \underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=0 \\ \sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{7n+3}}}{2{{n}^{2}}+4n+16}} \\ \text{Here }{{b}_{n}}=\dfrac{1}{2{{n}^{2}}+4n+16} \\ \text{Clearly }{{b}_{n}}\text{ is decreasing sequence } \\ \text{Also }\underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{2{{n}^{2}}+4n+16}=0 \\ \text{Hence the given series is convergent.}$

 What is the third multiple of two times of $\sum\limits_{n=1}^{\infty }{\dfrac{3}{{{7}^{n}}}}$

 First simplify the series and we see that it forms an infinite G.P with common ratio $\dfrac{1}{7}$, hence use the sum formula for a G.P and apply the given condition.

  $ \sum\limits_{n=1}^{\infty }{\dfrac{3}{{{7}^{n}}}} \\ =3\left\{ \dfrac{1}{7}+\dfrac{1}{{{7}^{2}}}+\dfrac{1}{{{7}^{3}}}+...... \right\} \\ \text{Sum of infinite G}\text{.P =}\dfrac{a}{1-r} \\ a=\dfrac{1}{7},r=\dfrac{1}{7} \\ \text{Sum =  3}\times \dfrac{\dfrac{1}{7}}{1-\dfrac{1}{7}} \\ =3\times \dfrac{1}{6}=\dfrac{1}{2} \\ 3\times 2\times \dfrac{1}{2}=3 $

 Test the convergence of the series $\sum\limits_{n=0}^{\infty }{\dfrac{5{{n}^{2}}+6n}{\left( n+1 \right)!}}$

 Check the value of $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|}$ and see whether the given series converges or diverges by checking the value whether finite or infinite.

  $ \sum\limits_{n=0}^{\infty }{\dfrac{5{{n}^{2}}+6n}{\left( n+1 \right)!}} \\ \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left| {{a}_{n+1}} \right|}{\left| {{a}_{n}} \right|} \\ =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{5{{\left( n+1 \right)}^{2}}+6\left( n+1 \right)}{\left( n+2 \right)!}\times \dfrac{\left( n+1 \right)!}{5{{n}^{2}}+6n} \\ =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\left( n+1 \right)\left( 5n+1 \right)}{\left( n+2 \right)n\left( 5n+6 \right)}=0 \\ \therefore \text{Convergent} $