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Section 1: Functions of Several Variables
Section 2: Limits and Continuity
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Practice Problems for Module 10
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.
Determine if the series converges or diverges.
\(\displaystyle \sum_{n=1}^\infty \frac{5\sin\left(\left(n+\frac{3}{2}\right)\pi\right)}{9+\sqrt{n}}\)
Answer
Converges
Video
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.
To see the full video page and find related videos, click the following link.
WIR 20B M152 V62
\(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n(n+2)}{3n^2+4n}\)
Answer
Converges
Video
To see the full video page and find related videos, click the following link.
WIR 20B M152 V63
\(\displaystyle \sum_{n=1}^\infty \frac{(-1)^{2n+1}\cos(n\pi)}{\sqrt[3]{n}}\)
Answer
Converges
Video
To see the full video page and find related videos, click the following link.
WIR 20B M152 V64
\(\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n^2}{\sqrt{n^4+2}}\)
Answer
Diverges
Video
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WIR 20B M152 V65
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.
Answer
The error is bounded by \( \dfrac{1}{e^5\cdot 5!} \approx 0.000056\).
Video
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WIR 20B M152 V66
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?
Answer
It would take at least \(100000^5-4\) terms.
Video
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WIR 20B M152 V67
Determine if the series is absolutely convergent, conditionally convergent, or divergent.
\(\displaystyle \sum_{n=2}^\infty \frac{(-1)^{n+1}}{n^{7/8}}\)
Answer
Conditionally Convergent
Video
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WIR 20B M152 V68
\(\displaystyle \sum_{n=2}^\infty \frac{n\cos(\ln(n))}{2n^3-8}\)
Answer
Absolutely Convergent
Video
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WIR 20B M152 V69
\(\displaystyle \sum_{n=1}^\infty \frac{(-1)^ne^{1/(n+1)}}{\sqrt[3]{n}}\)
Answer
Conditionally Convergent
Video
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WIR 20B M152 V70
\(\displaystyle \sum_{n=1}^\infty \frac{(-3)^{n-1}n^5}{4^{2n+1}}\)
Answer
Absolutely Convergent
Video
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WIR 20B M152 V71
\(\displaystyle \sum_{n=1}^\infty \frac{(-5)^nn!}{3\cdot7\cdot11\cdots(4n-1)}\)
Answer
Divergent
Video
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WIR 20B M152 V72
Find the interval and the radius of convergence for the following power series.
\(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n(x+5)^n}{n6^n}\)
Answer
Radius of Convergence: \(R=6\)
Interval of Convergence: \(I=\left(-11,1\right]\)
Video
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WIR 20B M152 V73
\(\displaystyle \sum_{n=1}^\infty \frac{n!(7x-3)^n}{9^n}\)
Answer
Radius of Convergence: \(R=0\)
Interval of Convergence: \(I=\left\{\dfrac{3}{7}\right\}\) or \(I=\left[\dfrac{3}{7},\dfrac{3}{7}\right]\)
Video
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WIR 20B M152 V74
\(\displaystyle \sum_{n=2}^\infty \frac{(-1)^n(2x-1)^n}{\ln(n)}\)
Answer
Radius of Convergence: \(R=\dfrac{1}{2}\)
Interval of Convergence: \(I=(0,1]\)
Video
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