# Practice Problems for Module 9

Section 11.4

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. Use the Comparison Test to determine if the series converges or diverges.
1. $$\displaystyle \sum_{n=5}^\infty \frac{7^n-8}{9+5^n}$$

Diverges
Video Errata: The presenter should have shown that 9 + 5n is less than something (not greater than) since we are dividing by it. Rather than $$9 + 5n ≥ 5n$$, he should have shown $$9+5n ≤2·5n$$. This follows from $$9<5^2 ≤ 5^n$$ for $$n≥2$$ and so $$9+5^n <5^n +5^n =2·5^n$$. Note the final answer is the same in that the series diverges, but with $$1/4$$ in front instead of $$1/2$$.

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2. $$\displaystyle \sum_{n=1}^\infty \frac{\ln(n)}{\sqrt{n}}$$ ​

Diverges

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

Converges
Video Errata: The $$\cos(1)$$ should be $$\cos(n)$$. The argument still works.

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4. $$\displaystyle \sum_{n=1}^\infty \frac{2n^3e^{-5n}+1}{7n^6+3n}$$

Converges

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2. ​Use the Limit Comparison Test to determine if the series converges or diverges.
1. $$\displaystyle \sum_{n=2}^\infty \frac{9n^3+2}{7n^4-\sin(5n)}$$

Diverges

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2.  $$\displaystyle \sum_{n=3}^\infty \frac{\sqrt{5n-6}}{3n^2+1}$$

Converges

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3. $$\displaystyle \sum_{n=2}^\infty \frac{7n^6+n^4}{\sqrt[3]{5n^2-1+n^{15}}}$$

Diverges

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