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This is one of our many quizzes on Convergence and divergence. Click on this to access more such competitive quizzes on Algebra.

Convergence and divergence Quiz 2

mathematics

 Which of the following series will converge for sure? 

 Correlate the given series with the general properties of each series to determine if they will converge or diverge in the given condition.

  $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ is a general summation series with the next term greater than the previous one since addition of two positive number will always result in a greater number. $ \\ $ $\sum\limits_{n=1}^{\infty }{a.{{r}^{n-1}}\text{ }\forall \text{ }r\le 1\text{ or }r\ge 1}$ shows a geometric series where the common ration will limit the next term in the given range of r, hence the series will diverge. $ \\ $ $\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{p}}}\text{ }\forall \text{p}\le \text{1}}$ represents a p-series and for the given range of p the value will always be a decimal lesser than 0 and the series will diverge.

 A student is stuck while solving for the next term of an equation where its general term is ${{\left( \dfrac{3n+1}{4-2n} \right)}^{2n}}$. He has the value of ‘n’ from 1 to Infinity. What will be the equation’s tendency with the axis?

 Check the conditions to apply the Root test for the limits. The series can be written as $\sum\limits_{n=1}^{\infty }{{{\left( \dfrac{3n+1}{4-2n} \right)}^{2n}}}$.

  Using the root test we have $ \\ L=\underset{n\to \infty }{\mathop{\lim }}\,{{\sqrt{|{{a}_{n}}|}}^{\frac{1}{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,{{\left| {{\left( \dfrac{3n+1}{4-2n} \right)}^{2n}} \right|}^{\frac{1}{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\left| {{\left( \dfrac{3n+1}{4-2n} \right)}^{2}} \right|={{\left| \left( \dfrac{-3}{2} \right) \right|}^{2}}=\dfrac{9}{4} \\ $ Since $\dfrac{9}{4}  >  1$, hence the series diverges.

 Check the nature of the equation $\sum\limits_{n=0}^{\infty }{\dfrac{1}{{{3}^{n}}}}$ with respect to the axis.

 Check if integration can be performed In order to use the Integral Test, integrate the equation $\sum\limits_{n=0}^{\infty }{\dfrac{1}{{{3}^{n}}}}$

$\sum\limits_{n=0}^{\infty }{\dfrac{1}{{{3}^{n}}}=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}} \\ $  Form the original series, writing down the partial sums $ \\ {{s}_{n}}=\sum\limits_{i=0}^{n}{\dfrac{1}{{{3}^{i}}}} \\ $  As seen, all the terms are positive, so it is an increasing sequence. $ \\ $ Since all the partial sums of the series are a convergent sequence therefore the series itself is also convergent.

 A physicists discovers a new particle which moves according to the path followed by the equation $\sum\limits_{n=2}^{\infty }{\dfrac{7}{n(n+1)}}$. What can you predict about its motion? 

 The $+1$ won't affect the equation much for a large value of n, so it is neglected and the series becomes $\dfrac{7}{{{n}^{2}}}$.

 The general term of the equation is ${{a}_{n}}=\dfrac{7}{n(n+1)}$ Here we can see that the series is positive for all terms so we can use a limit comparison test for this equation. The $+1$ will not affect the equation for a larger value of n so neglecting it.  $ \\ {{b}_{n}}=\dfrac{7}{n(n)}=\dfrac{7}{{{n}^{2}}} \\ $ We also know that the series  $\sum\limits_{n=2}^{\infty }{\dfrac{7}{{{n}^{2}}}}$ will converge by the p-series test $p=2 > 1$  $ \\ $ Therefore it can be thought that the parent equation will converge as well.  $ \\ $ For confirmation makes the numerator smaller or the denominator larger. $ \\ n < n+1 \\   n(n) < n(n+1) \\   \dfrac{7}{n(n)} > \dfrac{7}{n(n+1)} \\ $ By comparison test the parent series must also converge.

 Shalini has a progression such that its general term is   $\dfrac{{{n}^{2}}}{{{n}^{3}}-3}$ The initial value of the variable is 4 and the end value tends to infinity. Will the next term of the progression be smaller or larger than the previous? 

 P-series is not a confirmation test. Compare the fractions to find if a series converges or diverges.

 From the general term we get  $ \\ {{a}_{n}}=\dfrac{{{n}^{2}}}{{{n}^{3}}-3} \\ $  As long as $n > \sqrt[3]{3}$  Comparison test can be carried out. The  $-3$ is neglected  $ \\ {{b}_{n}}=\dfrac{{{n}^{2}}}{{{n}^{3}}}=\dfrac{1}{n} \\ \sum\limits_{n=4}^{\infty }{\dfrac{1}{n}} \\ $ Since it is a harmonic series tending to infinity, assumption can be made that it is diverging using the p-series test. $ \\ $ Here both numerator and denominator in the term,  ${{a}_{n}}=\dfrac{{{n}^{2}}}{{{n}^{3}}-3}$  are rational expressions, the term can be made smaller by either making the numerator smaller or denominator larger. $ \\ $ So by limit comparison test $ \\ {{n}^{3}} > {{n}^{3}}-3 \\ $ So, $\dfrac{{{n}^{2}}}{{{n}^{3}}} < \dfrac{{{n}^{2}}}{{{n}^{3}}-3}$  $ \\ $ The numerator in each term is the same while the denominator in the LHS is larger than RHS. Therefore the rational expression on LHS is smaller than the RHS.