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State, for each series, whether it convergence absolutely, convergence conditionally or...

Question:

State, for each series, whether it convergence absolutely, convergence conditionally or divergent. Justify the answer.

{eq}\sum _ { n = 2 } ^ { \infty } \frac { \operatorname { \cos } ( n ) } { n ( \ln (n)) ^ { 2 } }. {/eq}

Comparison Test:

If the infinite series {eq}\sum_{n=1}^{\infty} b_n {/eq} converges and {eq}0 \le a_n \le b_n {/eq} for all sufficiently large {eq}n {/eq} (that is, for all {eq}n>N {/eq} for some positive integer {eq}N {/eq}), then the infinite series {eq}\sum_{n=1}^{\infty} a_n {/eq} also converges. If the infinite series {eq}\sum_{n=1}^{\infty} b_n {/eq} diverges and {eq}0 \le b_n \le a_n {/eq} for all sufficiently large {eq}n {/eq}, then the infinite series {eq}\sum_{n=1}^{\infty} a_n {/eq} also diverges.


{eq}\displaystyle \ln {/eq} {eq}\displaystyle p {/eq} -Series: A series of the form {eq}\displaystyle \sum\limits_{n = 2}^\infty {\frac{1}{{n{{\left( {\ln \left( n \right)} \right)}^p}}}} {/eq} is called a {eq}\displaystyle \ln {/eq} -{eq}\displaystyle p {/eq} series, where {eq}\displaystyle p {/eq} is any real number.

A {eq}\displaystyle \ln {/eq} {eq}\displaystyle p {/eq} -series is convergent for {eq}\displaystyle p>1 {/eq} and divergent for {eq}\displaystyle p\leq 1. {/eq}


Answer and Explanation:

Let {eq}\displaystyle {a_n} = \frac{{ \cos \left( n \right)}}{{n{{\left( {\ln \left( n \right)} \right)}^2}}},\,\,\,n \geqslant 2. {/eq}


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Testing for Convergence & Divergence by Comparing Series

from AP Calculus BC: Exam Prep

Chapter 21 / Lesson 5
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