The interval of convergence of the series is [ 0 , 2 ] . Use this to find the sum of the series ...

Question:

The interval of convergence of the series is {eq}[0,2]{/eq}. Use this to find the sum of the series

{eq}\displaystyle \sum_{n=0}^\infty \frac{(-1)^n(n+1)}{2^n}{/eq}

(Hint: This is the same as {eq}\displaystyle \sum_{n=0}^\infty (-1)^n(n+1)\bigg (\frac{1}{2^n}\bigg ){/eq} so what got plugged in for {eq}x{/eq}?)

Sum of Power series:


If {eq}\sum\limits_{k = 0}^\infty {{a_k}} {/eq} is an infinite power series and {eq}{S_n} = \sum\limits_{k = 0}^n {{a_k}} {/eq} is the {eq}{n^{th}}{/eq} partial sum of series then {eq}\mathop {\lim }\limits_{n \to \infty } {S_n} = S{/eq} is tends to a real limit, this limit is called the sum of series.


Answer and Explanation: 1

Given:


  • The given series is {eq}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}\left( {n + 1} \right)}}{{{2^n}}}} {/eq} and interval of convergence is {eq}\left[ {0,2} \right] {/eq} .
  • The objective is to find the sum of series.


Rewrite the given series as:


{eq}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}\left( {n + 1} \right)}}{{{2^n}}}} = \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)\left( {n + 1} \right)} {/eq}


Split the series as follows:


{eq}\sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)\left( {n + 1} \right)} = \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)n + \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)} } {/eq}


Recall the result as follows:


{eq}\sum\limits_{n = 0}^\infty {n{z_0} = \dfrac{{{z_0}}}{{{{\left( {{z_0} - 1} \right)}^2}}}} {/eq}


Put {eq}{z_0} = - \dfrac{1}{2}{/eq}


{eq}\begin{align*} \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)n = \dfrac{{\dfrac{1}{2}}}{{{{\left( {\dfrac{{ - 1}}{2} - 1} \right)}^2}}}} \\ {\rm{ }} = - \dfrac{2}{9} \end{align*}{/eq}


{eq}\sum\limits_{n = 0}^\infty {\left( { - \dfrac{1}{{{2^n}}}} \right)} {/eq} is an infinite geometric series with {eq}b = 1,q = - \dfrac{1}{2}{/eq}.


Sum of the series {eq}S = \dfrac{b}{{1 - q}} = \dfrac{2}{3}{/eq}


Therefore the sumo of given series as follows:


{eq}\begin{align*} \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)\left( {n + 1} \right)}& = \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)n + \sum\limits_{n = 0}^\infty {\left( {\dfrac{{ - 1}}{{{2^n}}}} \right)} } \\ & = \dfrac{{ - 2}}{9} + \dfrac{2}{3}\\ & = \dfrac{4}{9}\\ & \approx 0.4444 \end{align*}{/eq}


Thus, the required sum of series is {eq}0.4444{/eq}.


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Power Series in X & the Interval of Convergence

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Chapter 12 / Lesson 6
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A power series in 'x' involves factors where an 'X' is added to a constant, and raised to a power, forming infinite terms. Learn how to build a power series and explore how the summing of these terms and a ratio test identifies the interval of convergence.


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