 # Practice Problems for Module 10

Sections 11.5, 11.6, and 11.8

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

Converges

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

Converges

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

Diverges

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

Absolutely Convergent

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

Conditionally Convergent

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

Absolutely Convergent

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

Divergent

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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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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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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.

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