# 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.

To see the full video page and find related videos, click the following link.

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

Converges

To see the full video page and find related videos, click the following link.

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

Converges

To see the full video page and find related videos, click the following link.

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

Diverges

To see the full video page and find related videos, click the following link.

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 $$\dfrac{1}{e^5\cdot 5!} \approx 0.000056$$.

To see the full video page and find related videos, click the following link.

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.

To see the full video page and find related videos, click the following link.

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

To see the full video page and find related videos, click the following link.

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

Absolutely Convergent

To see the full video page and find related videos, click the following link.

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

Conditionally Convergent

To see the full video page and find related videos, click the following link.

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

Absolutely Convergent

To see the full video page and find related videos, click the following link.

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

Divergent

To see the full video page and find related videos, click the following link.

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: $$I=\left(-11,1\right]$$

To see the full video page and find related videos, click the following link.

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 $$I=\left[\dfrac{3}{7},\dfrac{3}{7}\right]$$

To see the full video page and find related videos, click the following link.

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.