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

Root test Quiz 2

mathematics

 For a series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ with positive terms, if the value of $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}$ is greater than one, then series 

 The root test states that if $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}$ is greater than one the series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ diverges and if $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}$ is less than one then $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ converges.

  $\text{For a series} \sum\limits_{n=1}^{\infty }{{{a}_{n}}} \\ \text{With positive terms, if the value of} \\ \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}} \\ \text{If it is greater than one, then series will diverge}\text{.}$

 Which of the following statements is correct regarding root test?

 A series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ will be ready for root test if it has only positive terms.

  $\text{For a series} \\ \sum\limits_{n=1}^{\infty }{{{a}_{n}}}\\ \text{With positive terms, if the value of} \\ \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}} \\ \text{It is less than one, then series will converge}\text{.}$

 The series $\sum\limits_{n=0}^{\infty }{\dfrac{{{n}^{1-272n}}}{{{16}^{2n}}}}$ will

 For a series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ with positive terms, if the value of $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}$ is less than one, then the series will converge.

The nth term in the series $ \\ \sum\limits_{n=0}^{\infty}{\dfrac{n^{1-272n}}{16^{2n}}} \text{ is } a_n=\dfrac{n^{1-272n}}{16^{2n}} \\ $ By using root test ,  $ \\ \Rightarrow l=\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\dfrac{{{n}^{1-272n}}}{{{16}^{2n}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{n}^{\frac{1}{n}-272}}}{{{16}^{2}}}=0 \\ $ As the value of $l$ is less than one $ \\ $ The series will converge.

 The series $\sum\limits_{n=0}^{\infty }{{{\sin }^{n}}}\dfrac{\pi }{2}$ will 

 According to root test, for a series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ with positive terms, if the value of $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}$ is equal to one, then the series may be convergent or divergent.

 $\text{The nth term in the series} \\ \sum\limits_{n=0}^{\infty }{{{\sin }^{n}}}\dfrac{\pi }{2} \text{is} {{a}_{n}}={{\sin }^{n}}\dfrac{\pi }{2} \\ \text{By using root test ,} \\ l=\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{\sin }^{n}}\dfrac{\pi }{2}}=1 \\ \text{As the value of }l\text{ is equal to one} \\ \text{The series may be convergent or divergent}\text{.}$

 The series $\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1000)}^{n}}}{n}}$ will

 According to root test, for a series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ with positive terms, if the value of $\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{{{a}_{n}}}$ is greater than one, then the series is convergent.

 $\text{The } {n^{th}} \text{ term in the series} \\ \sum\limits_{n=0}^{\infty }{\dfrac{{{(-1000)}^{n}}}{n}} \text{is} {{a}_{n}}=\dfrac{{{(-1000)}^{n}}}{n} \\ \text{By using root test ,} \\ l=\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\dfrac{{{(-1000)}^{n}}}{n}}=-1000 \\ \text{As the value of }l\text{ is less than one} \\ \text{The series is convergent}$