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Practice Problems for Module 10

Sections 11.5, 11.6, and 11.8

Directions. The following are review problems. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. You can also follow the link for the full video page to find related videos. 
  1. Determine if the series converges or diverges.
    1. \(\displaystyle \sum_{n=1}^\infty \frac{5\sin\left(\left(n+\frac{3}{2}\right)\pi\right)}{9+\sqrt{n}}\) 

      Converges
      Video Errata: The presenter did not check that \(b_n\geq 0\), which you must do to apply the Alternating Series Test. In this case, it is obvious since \(\sqrt{n}\geq 0\), but it should be stated.
       

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      WIR 20B M152 V62


    2. \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n(n+2)}{3n^2+4n}\) 

      Converges

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      WIR 20B M152 V63


    3. \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^{2n+1}\cos(n\pi)}{\sqrt[3]{n}}\) 

      Converges

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      WIR 20B M152 V64


    4. \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n^2}{\sqrt{n^4+2}}\) 

      Diverges

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      WIR 20B M152 V65


  2. Consider the series: \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{e^nn!}\). If the sum of the series is estimated by the 4\(^\textrm{th}\) partial sum, then find the maximum error in this estimate. 

    The error is bounded by \( \dfrac{1}{e^5\cdot 5!} \approx 0.000056\).

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    WIR 20B M152 V66


  3. How many terms of the series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{(n+3)^{1/5}}\) are needed to find a partial sum that approximates the sum of the series to within \(0.0001\) accuracy? 

    It would take at least \(100000^5-4\) terms. 

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    WIR 20B M152 V67


  4. Determine if the series is absolutely convergent, conditionally convergent, or divergent. 
    1. \(\displaystyle \sum_{n=2}^\infty \frac{(-1)^{n+1}}{n^{7/8}}\)

      Conditionally Convergent

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      WIR 20B M152 V68


    2. \(\displaystyle \sum_{n=2}^\infty \frac{n\cos(\ln(n))}{2n^3-8}\)

      Absolutely Convergent

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      WIR 20B M152 V69


    3. \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^ne^{1/(n+1)}}{\sqrt[3]{n}}\)  

      Conditionally Convergent

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      WIR 20B M152 V70


    4. \(\displaystyle \sum_{n=1}^\infty \frac{(-3)^{n-1}n^5}{4^{2n+1}}\)

      Absolutely Convergent


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      WIR 20B M152 V71


    5. \(\displaystyle \sum_{n=1}^\infty \frac{(-5)^nn!}{3\cdot7\cdot11\cdots(4n-1)}\)

      Divergent


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      WIR 20B M152 V72


  5. Find the interval and the radius of convergence for the following power series.
    1. \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n(x+5)^n}{n6^n}\)

      Radius of Convergence: \(R=6\)
      Interval of Convergence: \(I=\left(-11,1\right]\)


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      WIR 20B M152 V73


    2. \(\displaystyle \sum_{n=1}^\infty \frac{n!(7x-3)^n}{9^n}\)

      Radius of Convergence: \(R=0\)
      Interval of Convergence: \(I=\left\{\dfrac{3}{7}\right\}\) or \(I=\left[\dfrac{3}{7},\dfrac{3}{7}\right]\)


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      WIR 20B M152 V74


    3. \(\displaystyle \sum_{n=2}^\infty \frac{(-1)^n(2x-1)^n}{\ln(n)}\)

      Radius of Convergence: \(R=\dfrac{1}{2}\)
      Interval of Convergence: \(I=(0,1]\)
      Video Errata: In the video, the series should start at \(n=2\). When \(n=1\), the denominator is \(\ln(1)\) which equals \(0\). The radius and interval of convergence in the video are correct. 
       


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V75