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

Limit comparison test Quiz 2

mathematics

 Chaitanya wrote a progression with the general term as $\dfrac{{{n}^{2}}}{{{n}^{3}}-3}$. The initial value of n is 4 and goes till infinity. He wonders if the next terms will be greater than the previous terms or smaller.

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

  Comparison test states that For two series  $\sum {{a}_{n}}$  and  $\sum {{b}_{n}}$  with $\sum {{a}_{n}}$ , $\sum {{b}_{n}} \geq 0$  for all $n$  and ${{a}_{n}}\le {{b}_{n}} \forall n$ Then, $ \\ $  If $\sum {{b}_{n}}$ is convergent then so is  $\sum {{a}_{n}}$. $ \\ $ If $\sum {{a}_{n}}$ is divergent then so is $\sum {{b}_{n}}$. $ \\ $ 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 $-3$ denominator will not affect the numerator when n tends to infinity and hence is neglected  $\\ {{b}_{n}}=\dfrac{{{n}^{2}}}{{{n}^{3}}}=\dfrac{1}{n} \\ $  Here, $\sum\limits_{n=4}^{\infty }{\dfrac{1}{n}}$  $ \\ $ Since it is a harmonic series tending to infinity, assumption can be made that it is a diverging series using p-series test. $ \\ $ For confirmation we prove that a series with smaller terms diverges so in turn a series with larger terms also diverges. $ \\ $ Since both the 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. $ \\ $ Also,  $ \\ {{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 the RHS. Therefore the rational expression on LHS is smaller than the RHS.

 A Mathematician is stuck with proving a theorem with the term $\dfrac{{{e}^{-n}}}{{{n}^{2}}+2n}$. She has the value of a variable from 0 to a value tending to $\infty $. What can you tell about its graph?

 Check the conditions to apply the Comparison test for the limits. The series can be written as $\sum\limits_{n=3}^{\infty }{\dfrac{{{e}^{-n}}}{{{n}^{2}}+2n}}$.

  Comparison test states that For two series  $\sum {{a}_{n}}$  and $\sum {{b}_{n}}$  with $\sum {{a}_{n}}$ , $\sum {{b}_{n}} \geq 0$ for all $n$  and ${{a}_{n}}\le {{b}_{n}}\forall n$ Then, $ \\ $ If $\sum {{b}_{n}}$ is convergent then so is $\sum {{a}_{n}}$. $ \\ $ If $\sum {{a}_{n}}$ is divergent then so is $\sum {{b}_{n}}$. $ \\ $ The series can be written as $\sum\limits_{n=3}^{\infty }{\dfrac{{{e}^{-n}}}{{{n}^{2}}+2n}}$ $ \\ $ For this range of n, the value is positive so Comparison Test can be applied $ \\ $ Since the numerator is exponential, the range of the numerator will turn towards 0 as n reaches to infinity. $ \\ $ The denominator is a polynomial. Since the exponent increases its value more than polynomial for a small increase in value of the determinant variable. $ \\ $ The numerator will reach 0 before the polynomial increases. $ \\ $ By this fact the series will probably converge $ \\ $ For confirmation, considering the numerator $ \\ {{e}^{-n}}<{{e}^{-3}}<1 \\ $  Since both the numerator and denominator in the term ${{a}_{n}}=\dfrac{{{e}^{-n}}}{{{n}^{2}}+2n}$ are rational expressions, the term can be made smaller by either making the numerator smaller or denominator larger. $ \\ $ We can make the numerator larger by the above relationship $ \\ \dfrac{{{e}^{-n}}}{{{n}^{2}}+2n}<\dfrac{1}{{{n}^{2}}+2n} \\ $  Now, in the denominator it is seen that, $ \\ {{n}^{2}}+2n>{{n}^{2}} \\ $  Using this we can make the denominator smaller, thereby making the expression larger $ \\ \dfrac{{{e}^{-n}}}{{{n}^{2}}+2n}<\dfrac{1}{{{n}^{2}}+2n}<~\dfrac{1}{{{n}^{2}}} \\ $ With P-series test and comparison of numerators, it is clarified that the series must converge. 

 The equation $\sum\limits_{n=0}^{\infty }{\dfrac{{{2}^{n}}\sin (5n)}{{{4}^{n}}+{{\cos }^{2}}(n)}}$  is used in deep learning and neural networks for artificial intelligence. Check if more nodes are required or not to store information.

 Check the conditions to apply the Comparison test for the limits. The general term can be written as ${{a}_{n}}=\dfrac{{{2}^{n}}\sin (5n)}{{{4}^{n}}+{{\cos }^{2}}(n)}$.

Comparison test states that For two series $\sum {{a}_{n}}$  and $\sum {{b}_{n}}$  with $\sum {{a}_{n}}$ , $\sum {{b}_{n}} \geq 0$  for all $n$  and ${{a}_{n}}\le {{b}_{n}}\forall n$ Then, $ \\ $ If $\sum {{b}_{n}}$ is convergent then so is $\sum {{a}_{n}}$. $ \\ $ If $\sum {{a}_{n}}$ is divergent then so is $\sum {{b}_{n}}$. $ \\ $ The General can be written as ${{a}_{n}}=\dfrac{{{2}^{n}}\sin (5n)}{{{4}^{n}}+{{\cos }^{2}}(n)}$ $ \\ $ For this range of n, the value is positive so Comparison Test can be applied $ \\ $ Since the numerator and denominator contains Trigonometric function, they won’t affect the size of the numerator and denominator for large values of $n$   $ \\ $ So for large $n$ the term will behave like, $ \\ {{b}_{n}}=\dfrac{{{2}^{n}}}{{{4}^{n}}}={{\left( \dfrac{2}{4} \right)}^{n}}={{\left( \dfrac{1}{2} \right)}^{n}} \\ $  We know that the series $\sum\limits_{n=0}^{\infty }{{{\left( \dfrac{1}{2} \right)}^{n}}}$ will converted because it is a Geometric Series with $r=\dfrac{1}{2}<1$  $ \\ $ By this fact the series will probably converge $ \\ $ For confirmation, considering the numerator $ \\ $ We know that $0\le {{\sin }^{2}}(5n)\le 1$  $ \\ $ So replacing the ${{\sin }^{2}}(5n)$ in the numerator with the largest possible value we get, $ \\ {{2}^{n}}{{\sin }^{2}}(5n)\le {{2}^{n}}1={{2}^{n}} \\ $ Using this we can make the numerator larger to get the following relationship $ \\ \dfrac{{{2}^{n}}{{\sin }^{2}}(5n)}{{{4}^{n}}+{{\cos }^{2}}(n)}\le \dfrac{{{2}^{n}}}{{{4}^{n}}+{{\cos }^{2}}(n)} \\ $ Now in denominator we know that $0\le {{\cos }^{2}}(n)\le 1$ and so replacing the ${{\cos }^{2}}(n)$with the smallest possible value we get, $ \\ {{4}^{n}}+{{\cos }^{2}}(n)>{{4}^{n}}+0={{4}^{n}} \\ $ Using this make the denominator smaller, and thereby making the expression larger to get $ \\ \dfrac{{{2}^{n}}{{\sin }^{2}}(5n)}{{{4}^{n}}+{{\cos }^{2}}(n)}\le \dfrac{{{2}^{n}}}{{{4}^{n}}+{{\cos }^{2}}(n)}\le \dfrac{{{2}^{n}}}{{{4}^{n}}}={{\left( \dfrac{1}{2} \right)}^{n}} \\ $ This is a convergent series Therefore by Comparison test the series given must also converge.

 A gaming company wishes to make a visor that moves with the character’s vision. For this they need to check the nature of the equation$\sum\limits_{n=0}^{\infty }{\dfrac{1}{{{3}^{n}}+n}}$. 

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

 Comparison test states that For two series $\sum {{a}_{n}}$  and $\sum {{b}_{n}}$  with $\sum {{a}_{n}}$, $\sum {{b}_{n}} \geq 0$ for all $n$  and ${{a}_{n}}\le {{b}_{n}}\forall n$ Then, $ \\ $ If $\sum {{b}_{n}}$ is convergent then so is $\sum {{a}_{n}}$. $ \\ $ If $\sum {{a}_{n}}$ is divergent then so is $\sum {{b}_{n}}$. $ \\ $ In order to use the Integral Test, integrate the equation $\sum\limits_{n=0}^{\infty }{\dfrac{1}{{{3}^{n}}+n}}$ $ \\ $ For integrating we write the equation as  $ \\ \int_{0}^{\infty }{\dfrac{1}{{{3}^{x}}+x}} \\ $  As seen, this equation cannot be integrated directly $ \\ $ So using alternate method $ \\ $ Regardless of the value of $x$ the value of ${{3}^{x}}>0$so we drop x from denominator as it can be neglected $ \\ \dfrac{1}{{{3}^{n}}+n}<\dfrac{1}{{{3}^{n}}}$ here $\sum\limits_{n=0}^{\infty }{\dfrac{1}{{{3}^{n}}}=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}}$   From original series writing down partial sums $ \\ {{s}_{n}}=\sum\limits_{i=0}^{n}{\dfrac{1}{{{3}^{i}}+1}} \\ $  As seen, all terms are positive so it is an increasing sequence. Since the partial sums of a series is a convergent sequence therefore the series itself is also convergent.

 A particle travels around a path that is represented 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 $\dfrac{7}{{{n}^{2}}}$.

 Comparison test states that For two series $\sum {{a}_{n}}$  and $\sum {{b}_{n}}$  with $\sum {{a}_{n}}$ , $\sum {{b}_{n}} \geq 0$ for all $n$  and ${{a}_{n}}\le {{b}_{n}} \forall n$ Then, $ \\ $ If $\sum {{b}_{n}}$ is convergent then so is $\sum {{a}_{n}}$. $ \\ $ If $\sum {{a}_{n}}$ is divergent then so is $\sum {{b}_{n}}$. $ \\ $ The general term of the equation is ${{a}_{n}}=\dfrac{7}{n(n+1)}$ $ \\ $ Here we see that the series is positive for all terms so we can use Limit comparison test for this equation $ \\ $ The $+1$ won't affect the equation much for a large value of n, so it is neglected $ \\ {{b}_{n}}=\dfrac{7}{n(n)}=\dfrac{7}{{{n}^{2}}} \\ $ We also know that 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 make 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, the parent series must also converge.