Solutions to Problem Set 19

Math 211-03

10-24-2017

[Direct comparison]

1. Use direct comparison to determine whether the series $\displaystyle \sum_{n=2}^\infty
   \dfrac{5}{n^3 + 3}$ converges or diverges.

$$\dfrac{5}{n^3 + 3} < \dfrac{5}{n^3}.$$

$\displaystyle \sum_{n=2}^\infty
   \dfrac{5}{n^3}$ converges, because it's 5 times a p-series with $p = 3 > 1$ . Therefore, $\displaystyle \sum_{n=2}^\infty
   \dfrac{5}{n^3 + 3}$ converges by direct comparison.


2. Use direct comparison to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{3}{n^{1/2} - 6}$ converges or diverges.

$$\dfrac{3}{n^{1/2} - 6} > \dfrac{3}{n^{1/2}}.$$

$\displaystyle \sum_{n=1}^\infty
   \dfrac{3}{n^{1/2}}$ diverges, because it's 3 times a p-series with $p = \dfrac{1}{2} < 1$ . Therefore, $\displaystyle
   \sum_{n=1}^\infty \dfrac{3}{n^{1/2} - 6}$ diverges by direct comparison.


3. Use direct comparison to determine whether the series $\displaystyle \sum_{n=5}^\infty
   \dfrac{2 n + 1}{n^{4/3} - 5}$ converges or diverges.

$$\dfrac{2 n + 1}{n^{4/3} - 5} > \dfrac{2 n}{n^{4/3} - 5} > \dfrac{2 n}{n^{4/3}} = \dfrac{2}{n^{1/3}}.$$

$\displaystyle \sum_{n=5}^\infty
   \dfrac{2}{n^{1/3}}$ diverges, because it's 2 times a p-series with $p = \dfrac{1}{3} < 1$ . Therefore, $\displaystyle
   \sum_{n=5}^\infty \dfrac{2 n + 1}{n^{4/3} - 5}$ diverges by direct comparison.


4. Use direct comparison to determine whether the series $\displaystyle \sum_{n=2}^\infty
   \dfrac{4^n - 5}{3 + 7^n}$ converges or diverges.

$$\dfrac{4^n - 5}{3 + 7^n} < \dfrac{4^n}{3 + 7^n} < \dfrac{4^n}{7^n} = \left(\dfrac{4}{7}\right)^n.$$

$\displaystyle \sum_{n=2}^\infty
   \left(\dfrac{4}{7}\right)^n$ converges, because it's a geometric series with ratio $r = \dfrac{4}{7}$ and $-1 < \dfrac{4}{7} < 1$ . Therefore, $\displaystyle
   \sum_{n=2}^\infty \dfrac{4^n - 5}{3 + 7^n}$ converges by direct comparison.


5. Use direct comparison to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \left(\dfrac{1}{n + 2}\right)^n$ converges or diverges.

$$\left(\dfrac{1}{n + 2}\right)^n < \left(\dfrac{1}{2}\right)^n.$$

$\displaystyle \sum_{n=1}^\infty
   \left(\dfrac{1}{2}\right)^n$ converges, because it's a geometric series with ratio $r = \dfrac{1}{2}$ and $-1 < \dfrac{1}{2} < 1$ . Therefore, $\displaystyle
   \sum_{n=1}^\infty \left(\dfrac{1}{n + 2}\right)^n$ converges by direct comparison.


6. Use direct comparison to determine whether the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{2 \cos n + 3}{n^2}$ converges or diverges.

$$-1 \le \cos n \le 1, \quad\hbox{so}\quad -2 \le 2 \cos n \le 2, \quad\hbox{and}\quad 1 \le 2 \cos n + 3 \le 5.$$

In particular, $2 \cos n + 3$ is positive, so the series has positive terms. The inequality above also shows that

$$\dfrac{2 \cos n + 3}{n^2} \le \dfrac{5}{n^2}.$$

$\displaystyle \sum_{n=1}^\infty
   \dfrac{5}{n^2}$ converges, because it's 5 times a p-series with $p = 2 > 1$ . Therefore, $\displaystyle \sum_{n=1}^\infty
   \dfrac{2 \cos n + 3}{n^2}$ converges by direct comparison.


7. Use direct comparison to determine whether the series $\displaystyle \sum_{n=2}^\infty
   \dfrac{n^{1/3}}{n - \sqrt{n}}$ converges or diverges.

$$\dfrac{n^{1/3}}{n - \sqrt{n}} > \dfrac{n^{1/3}}{n} = \dfrac{1}{n^{2/3}}.$$

$\displaystyle \sum_{n=2}^\infty
   \dfrac{1}{n^{2/3}}$ diverges, since it's a p-series with $p =
   \dfrac{2}{3} < 1$ . Therefore, $\displaystyle \sum_{n=2}^\infty
   \dfrac{n^{1/3}}{n - \sqrt{n}}$ diverges by direct comparison.


8. Use direct comparison to determine whether the series $\displaystyle \sum_{n=2}^\infty
   \dfrac{11^n + 3^n}{8^n - 5^n}$ converges or diverges.

$$\dfrac{11^n + 3^n}{8^n - 5^n} > \dfrac{11^n}{8^n - 5^n} > \dfrac{11^n}{8^n} = \left(\dfrac{11}{8}\right)^n.$$

$\displaystyle \sum_{n=2}^\infty
   \left(\dfrac{11}{8}\right)^n$ diverges, because it's a geometric series with ratio $r = \dfrac{11}{8}$ , and $\dfrac{11}{8} > 1$ . Therefore, $\displaystyle
   \sum_{n=2}^\infty \dfrac{11^n + 3^n}{8^n - 5^n}$ diverges by direct comparison.


All action is involved in imperfection, like fire in smoke. - The Bhagavad Gita


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