 # Practice Problems for Module 7

Covering Sections 11.1 and 11.2

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. Find a formula for the general term, $$a_n$$, of the sequence assuming that the pattern of the first few terms continues. Give the formula so that the first term is $$a_1$$.
$\frac{5}{27}, -\frac{9}{64}, \frac{13}{125}, -\frac{17}{216}, \ldots$

$$a_n=\dfrac{ (-1)^{n+1}(4n+1)}{(n+2)^3}$$

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2. Determine whether each sequence converges or diverges. If it converges, find the limit.
1. $$a_n=\dfrac{\ln(\ln(2n))}{5n}$$

Converges to 0

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2. $$a_n=\dfrac{(-1)^n n^3}{7n^3-1}$$

Diverges

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3. $$a_n=\arctan\left(\dfrac{1-n^2}{3n+1}\right)$$

Converges to $$-\dfrac{\pi}{2}$$

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3. ​Determine whether the following sequences are increasing, decreasing, or not monotonic. Also, determine if the sequences are bounded.
1. $$a_n=\ln(8+9n)-\ln(3n-2)$$

Bounded and Decreasing

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2. $$a_n=\tan\left( \dfrac{(2n-1)\pi}{4}\right)$$

Bounded and Not Monotonic
Video errata: The presenter spoke about convergence in the video rather than monotonicity. Not only does it not converge, but it is not monotonic. This is because no matter how far out we go in the sequence, it will still continue alternating between two fixed numbers, $$−1$$ and $$1$$.

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4. The sequence below is bounded and increasing (check!). Does it converge and if so, what to? $a_1=3 \qquad \qquad a_{n+1}=\dfrac{5a_n-4}{a_n}$

It converges to 4.

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5. Determine if the series converges or diverges. Give the sum if it converges.
1. $$\displaystyle \sum_{k=1}^\infty k^2 \sin\left( \frac{1}{k^2}\right)$$

Diverges
Video errata: The presenter said once towards the end of the video that the series converges due to the inside of the series converging to something not equal to zero, but should have said diverges.

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

Converges to $$-\dfrac{7}{8}$$

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3. $$\displaystyle \sum_{k=1}^\infty\left( e^{4/(k+1)}-e^{4/k}\right)$$

Converges to $$1-e^4$$

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