Solutions to Problem Set 25

Math 211-03

11-7-2017

[Interval of convergence]

1. Find the interval of convergence of the series $\displaystyle \sum_{n=0}^\infty
   \dfrac{5^n}{n!}(x + 3)^n$ .

$$\lim_{n \to \infty} \dfrac{\dfrac{5^{n + 1}}{(n + 1)!}|x + 3|^{n + 1}} {\dfrac{5^n}{n!}|x + 3|^n} = \lim_{n \to \infty} \dfrac{5^{n + 1}}{5^n} \cdot \dfrac{n!}{(n + 1)!} \cdot |x + 3| =$$

$$\lim_{n \to \infty} 5 \cdot \dfrac{1 \cdot 2 \cdots n}{1 \cdot 2 \cdots n \cdot (n + 1)} \cdot |x + 3| = \lim_{n \to \infty} \dfrac{5}{n + 1} |x + 3| = 0.$$

Since the limiting ratio is always less than 1, the series converges for all x. The interval of convergence is $(-\infty, \infty)$ .


2. Find the interval of convergence of the series $\displaystyle \sum_{n=1}^\infty
   \dfrac{(x - 2)^n}{n 5^n}$ .

$$\lim_{n \to \infty} \dfrac{\dfrac{1}{(n + 1) 5^{n + 1}}|x - 2|^{n + 1}}{\dfrac{1}{n 5^n}|x - 2|^n} = \lim_{n \to \infty} \dfrac{n}{n + 1} \cdot \dfrac{5^n}{5^{n + 1}} \cdot |x - 2| =$$

$$\lim_{n \to \infty} \dfrac{n}{n + 1} \cdot \dfrac{1}{5} |x - 2| = \dfrac{1}{5} |x - 2|.$$

The series converges absolutely for $\dfrac{1}{5} |x - 2| < 1$ . Solving for x, I get

$$|x - 2| < 5, \quad\hbox{or}\quad -3 < x < 7.$$

When $x = 7$ , the series is

$$\sum_{n=1}^\infty \dfrac{5^n}{n 5^n} = \sum_{n=1}^\infty \dfrac{1}{n}.$$

This is the harmonic series, so it diverges.

When $x = -3$ , the series is

$$\sum_{n=1}^\infty \dfrac{1}{n 5^n} \cdot (-5)^n = \sum_{n=1}^\infty \dfrac{(-1)^n}{n}.$$

This is the alternating harmonic series, so it converges.

The interval of convergence is $-3 \le x < 7$ .


3. Find the interval of convergence of the series $\displaystyle \sum_{n=0}^\infty
   \dfrac{n + 1}{5^n} x^n$ .

$$\lim_{n \to \infty} \dfrac{\dfrac{n + 2}{5^{n + 1}}|x|^{n + 1}}{\dfrac{n + 1}{5^n}|x|^n} = \lim_{n \to \infty} \dfrac{n + 2}{n + 1} \cdot \dfrac{5^n}{5^{n + 1}} \cdot |x| = \lim_{n \to \infty} \dfrac{n + 2}{n + 1} \cdot \dfrac{1}{5} |x| = \dfrac{1}{5} |x|.$$

The series converges absolutely for $\dfrac{1}{5} |x| < 1$ . Solving for x, I get

$$|x| < 5, \quad\hbox{or}\quad -5 < x < 5.$$

When $x = 5$ , the series is

$$\sum_{n=0}^\infty \dfrac{n}{5^n} \cdot 5^n = \sum_{n=0}^\infty n.$$

Since $\displaystyle \lim_{n \to
   \infty} n = +\infty \ne 0$ , the series diverges by the Zero Limit Test.

When $x = -5$ , the series is

$$\sum_{n=0}^\infty \dfrac{n}{5^n} \cdot (-5)^n = \sum_{n=0}^\infty (-1)^n n.$$

Since $\displaystyle \lim_{n \to
   \infty} (-1)^n n$ is undefined, the series diverges by the Zero Limit Test.

The interval of convergence is $-5 < x < 5$ .


4. Find the interval of convergence of the series $\displaystyle \sum_{n=0}^\infty n!
   (x - 4)^n$ .

$$\lim_{n \to \infty} \dfrac{(n + 1)! (x - 4)^{n + 1}}{n! |x - 4|^n} = \lim_{n \to \infty} \dfrac{(n + 1)!}{n!} \cdot \dfrac{|x - 4|^{n + 1}}{|x - 4|^n} = \lim_{n \to \infty} \dfrac{1 \cdot 2 \cdots n \cdot (n + 1)}{1 \cdot 2 \cdots n} \cdot |x - 4| =$$

$$\lim_{n \to \infty} (n + 1) |x - 4| = +\infty.$$

Since the limiting ratio is always greater than 1, the series only converges at the expansion point $x = 4$ .


5. Find the interval of convergence of the series $\displaystyle \sum_{n=0}^\infty
   \dfrac{(x - 1)^n}{(n + 1)^2}$ .

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

The series converges absolutely for $|x - 1| < 1$ . Solving for x, I get $0 < x < 2$ .

When $x = 2$ , the series is $\displaystyle \sum_{n=0}^\infty \dfrac{1}{(n + 1)^2}$ . This is a p-series with $p = 2 > 1$ , so it converges.

When $x = 0$ , the series is $\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{(n + 1)^2}$ . This is an alternating p-series with $p = 2 > 1$ , so it converges.

The interval of convergence is $0 \le x \le 2$ .


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