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

Alternating series test Quiz 2

mathematics

 Which alternating series convergence is true among the following.

 Apply leibniz’s test, leibniz’s rule for alternating series test. 

  $   \sum\limits_{n=1}^{\infty }{{{u}_{n}}}\text{ }with\text{ }both\text{ }positive\text{ }and\text{ }negative\text{ }term. \\   \text{A series of the form }\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}{{b}_{n}}}\text{ or }\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}{{b}_{n}}}\text{ where }{{\text{b}}_{n}}>0\text{ for all }n,\text{ is called an alternating} \\   \text{series, because the terms alternate between positive and negative value}\text{.} \\ $

 What is the value of absolute convergence series $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}.\dfrac{{{\left( n-1 \right)}^{3}}\left( n+1 \right)}{{{n}^{4}}+{{n}^{2}}}}$ at $n=7$.

 Apply absolute convergence series which is always positive.

  $   \text{Given }\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{{{\left( n-1 \right)}^{3}}\left( n+1 \right)}{{{n}^{4}}+{{n}^{2}}}} \\   \text{Where }n=7 \\   \Rightarrow \left| \dfrac{{{\left( -1 \right)}^{7}}{{\left( 7-1 \right)}^{3}}\left( 7+1 \right)}{{{\left( 7 \right)}^{4}}+{{\left( 7 \right)}^{2}}} \right| \\   \Rightarrow \left| \dfrac{{{\left( -1 \right)}^{7}}{{\left( 6 \right)}^{3}}\times 8}{2401+49} \right| \\   \Rightarrow \left| \dfrac{-1728}{2450} \right| \\   \Rightarrow 0.705 \\ $ 

 The following series diverges if $\sum\limits_{n=n}^{\infty }{{{\left( -1 \right)}^{n}}.\sec \left( \dfrac{9\pi }{2n} \right)}$ where $n=5$ is the term.

 Apply $\dfrac{\pi }{10}=18{}^\circ $ to calculate the value of $\sec \theta $.

$   \text{Given }\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\sec \left( \dfrac{9\pi }{2n} \right)\text{ at }n=5} \\   \therefore {{\left( -1 \right)}^{5}}\sec \left( \dfrac{9\pi }{2n} \right) \\   \Rightarrow {{\left( -1 \right)}^{5}}\sec \left( \pi -\dfrac{\pi }{10} \right) \\   \Rightarrow \sec \left( \dfrac{\pi }{10} \right) \\   \Rightarrow 1.0514 \\ $ 

 What is the partial sum of approximations of the sum of series $\sum\limits_{n=n}^{\infty }{{{\left( -1 \right)}^{n}}.\dfrac{1}{{{n}^{2}}\left( n+1 \right)}}$ where the error is less than $0.01={{10}^{-2}}$.

 Apply alternating series estimation theorem i.e., ${{b}_{n+1}}<{{b}_{n}}$ and $\underset{x\to \infty }{\mathop{\lim }}\,{{b}_{n}}=0$ then $\left| {{R}_{n}} \right|=\left| S-{{S}_{n}} \right|\le {{b}_{n+1}}$.

 $   \text{Given }\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{1}{{{n}^{2}}\left( n+1 \right)}>0\text{ for the n}\ge \text{1}} \\   \therefore \left| {{R}_{n}} \right|=\left| S-{{S}_{n}} \right|\le {{b}_{n+1}}=\dfrac{1}{{{\left( n+1 \right)}^{2}}\left( n+2 \right)} \\   \therefore \dfrac{1}{{{\left( n+1 \right)}^{2}}\left( n+2 \right)}\le \dfrac{1}{{{10}^{2}}} \\   \therefore {{R}_{n}}\le \dfrac{1}{{{10}^{2}}} \\   \Rightarrow \dfrac{1}{{{\left( n+1 \right)}^{2}}\left( n+2 \right)}\le \dfrac{1}{{{10}^{2}}}\text{ if }{{10}^{2}}\le {{\left( n+1 \right)}^{2}}\left( n+2 \right) \\   \Rightarrow 100\le \left( {{n}^{2}}+1+2n \right)\left( n+2 \right) \\   \Rightarrow 100\le {{n}^{3}}+n+2{{n}^{2}}+2{{n}^{2}}+2+4n \\   \Rightarrow 100\le {{n}^{3}}+4{{n}^{2}}+5n+2 \\   \Rightarrow 98\le n\left( {{n}^{2}}+4n+5 \right) \\ $

 What is the value of series for convergence $\sum\limits_{n=1}^{\infty }{\left( -1 \right)\dfrac{\log n}{n\left( n+1 \right)}}$, where $n=6$.

 Firstly simplify then apply the L'HOP rule to $\underset{x\to \infty }{\mathop{\lim }}\,$ in the function.

 $   \text{Given }\sum\limits_{n=1}^{\infty }{\left( -1 \right){{n}^{2}}\dfrac{\log n}{n+1}} \\   \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\log n}{n\left( n+1 \right)} \\   \text{Let }n=x \\   \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log x}{x\left( x+1 \right)} \\   \Rightarrow L'HOP\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{1}{x}}{x\left( x+1 \right)}\text{ (Apply L }\!\!'\!\!\text{ HOP rule)} \\   \therefore \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{x}^{3}}+x}=0 \\ $