Solutions to Problem Set 24

Math 211-03

11-6-2017

[Absolute convergence]

1. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{1}{(n + 2)!}$ converges absolutely, converges conditionally, or diverges.

The absolute value series is $\displaystyle \sum_{n=2}^\infty \dfrac{1}{(n + 2)!}$ . Apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{\dfrac{1}{(n + 3)!}}{\dfrac{1}{(n + 2)!}} = \lim_{n \to \infty} \dfrac{1}{(n + 3)!} \cdot \dfrac{(n + 2)!}{1} = \lim_{n \to \infty} \dfrac{(n + 2)!}{(n + 3)!} = \lim_{n \to \infty} \dfrac{1 \cdot 2 \cdots (n + 2)} {1 \cdot 2 \cdots (n + 2) \cdot (n + 3)} =$$

$$\lim_{n \to \infty} \dfrac{1}{n + 3} = 0 < 1.$$

The absolute value series converges. Therefore, the original series converges absolutely.


2. Determine whether the series $\displaystyle \sum_{n=0}^\infty (-1)^n 4^n$ converges absolutely, converges conditionally, or diverges.

$$\lim_{n \to \infty} 4^n = +\infty, \quad\hbox{so}\quad \lim_{n \to \infty} (-1)^n 4^n \quad\hbox{is undefined}.$$

The series diverges by the Zero Limit Test.


3. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{1}{\sqrt{2 n +
   3}}$ converges absolutely, converges conditionally, or diverges.

The absolute value series is $\displaystyle \sum_{n=2}^\infty \dfrac{1}{\sqrt{2 n + 3}}$ . Apply Limit Comparison:

$$\lim_{n \to \infty} \dfrac{\dfrac{1}{\sqrt{2 n + 3}}}{\dfrac{1}{\sqrt{n}}} = \lim_{n \to \infty} \dfrac{1}{\sqrt{2 n + 3}} \cdot \dfrac{\sqrt{n}}{1} = \lim_{n \to \infty} \dfrac{\sqrt{n}}{\sqrt{2 n + 3}} = \dfrac{1}{\sqrt{2}}.$$

The limit is a finite positive number. $\displaystyle \sum_{n=2}^\infty
   \dfrac{1}{\sqrt{n}}$ diverges, since it's a p-series with $p =
   \dfrac{1}{2} < 1$ . Therefore, the absolute value series diverges, and the original series does not converge absolutely.

Consider the original series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{1}{\sqrt{2 n +
   3}}$ .

The terms alternate, and

$$\lim_{n \to \infty} \dfrac{1}{\sqrt{2 n + 3}} = 0.$$

Let $f(x) = \dfrac{1}{\sqrt{2 x
   + 3}} = (2 x + 3){-1/2}$ . Then

$$f'(x) = -(2 x + 3)^{-3/2} < 0.$$

Hence, the terms decrease in absolute value.

Therefore, the original series converges by the Alternating Series Test.

Hence, the original series converges conditionally.


4. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^{n+1} \dfrac{3 n + 1}{4 n
   + 1}$ converges absolutely, converges conditionally, or diverges.

$$\lim_{n \to \infty} \dfrac{3 n + 1}{4 n + 1} = \dfrac{3}{4}, \quad\hbox{so}\quad \lim_{n \to \infty} (-1)^{n+1} \dfrac{3 n + 1}{4 n + 1} \quad\hbox{is undefined}.$$

The series diverges by the Zero Limit Test.


5. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{1}{(\sqrt{n} +
   1)^2}$ converges absolutely, converges conditionally, or diverges.

The absolute value series is $\displaystyle \sum_{n=2}^\infty \dfrac{1}{(\sqrt{n} + 1)^2}$ . Apply Limit Comparison:

$$\lim_{n \to \infty} \dfrac{\dfrac{1}{(\sqrt{n} + 1)^2}}{\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{1}{(\sqrt{n} + 1)^2} \cdot \dfrac{n}{1} = \lim_{n \to \infty} \dfrac{n}{n + 2 \sqrt{n} + 1} = 1.$$

The limit is a finite positive number. $\displaystyle \sum_{n=2}^\infty
   \dfrac{1}{n}$ diverges, since it's the harmonic series. Therefore, the absolute value series diverges, and the original series does not converge absolutely.

Consider the original series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{1}{(\sqrt{n} +
   1)^2}$ .

The terms alternate, and

$$\lim_{n \to \infty} \dfrac{1}{(\sqrt{n} + 1)^2} = 0.$$

Let $f(x) = \dfrac{1}{(\sqrt{x}
   + 1)^2} = (\sqrt{x} + 1)^{-2}$ . Then

$$f'(x) = -2 (\sqrt{x} + 1)^{-3} \left(\dfrac{1}{2 \sqrt{x}}\right) < 0.$$

Hence, the terms decrease in absolute value.

Therefore, the original series converges by the Alternating Series Test.

Hence, the original series converges conditionally.


6. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{n^2}{n^3 + 1}$ converges absolutely, converges conditionally, or diverges.

The absolute value series is $\displaystyle \sum_{n=1}^\infty \dfrac{n^2}{n^3 + 1}$ . Apply Limit Comparison:

$$\lim_{n \to \infty} \dfrac{\dfrac{n^2}{n^3 + 1}}{\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{n^2}{n^3 + 1} \cdot \dfrac{n}{1} = \lim_{n \to \infty} \dfrac{n^3}{n^3 + 1} = 1.$$

The limit is a finite positive number. $\displaystyle \sum_{n=2}^\infty
   \dfrac{1}{n}$ diverges, since it's the harmonic series. Therefore, the absolute value series diverges, and the original series does not converge absolutely.

Consider the original series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{n^2}{n^3 + 1}$ .

The terms alternate, and

$$\lim_{n \to \infty} \dfrac{n^2}{n^3 + 1} = 0.$$

Let $f(x) = \dfrac{x^2}{x^3 +
   1}$ . Then

$$f'(x) = \dfrac{(x^3 + 1)(2 x) - (x^2)(3 x^2)}{(x^3 + 1)^2} = \dfrac{2 x - x^4}{(x^3 + 1)^2} = \dfrac{x(2 - x^3)}{(x^3 + 1)^2}.$$

Now $2 - x^3 < 0$ for $x
   \ge 2$ , so $f'(x) < 0$ for $x \ge 2$ . Hence, the terms decrease in absolute value.

Therefore, the original series converges by the Alternating Series Test.

Hence, the original series converges conditionally.


7. The series $\displaystyle
   \sum_{n=1}^\infty \dfrac{\cos n}{2^n}$ has infinitely many positive and negative terms, so the convergence tests for positive term series don't apply.

Show that the series converges by showing that it converges absolutely.

The absolute value series is $\displaystyle \sum_{n=1}^\infty \dfrac{|\cos n|}{2^n}$ .

From trigonometry, $|\cos n| \le
   1$ . Hence,

$$\dfrac{|\cos n|}{2^n} \le \dfrac{1}{2^n}.$$

$\displaystyle \sum_{n=1}^\infty
   \dfrac{1}{2^n}$ converges, because it's a geometric series with ratio $\dfrac{1}{2}$ , and $-1 <
   \dfrac{1}{2} < 1$ . Therefore, the absolute value series converges by Direct Comparison, and the original series converges absolutely.

Therefore, the original series converges.


Adversity makes no impression upon a brave soul. - Ali Ibn Abi Talib


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga