Worked examples · Sheet 1

Radius and set of convergence with the ratio test

Theory used: Ratio test for the radius Radius of convergence theorem Set of convergence Study of the endpoints

Exercise 1

Determine the radius and the set of convergence of the series

n = 1 \sum \infty \frac{x ^{n}}{n 5 ^{n}} .

a_{n} = \frac{1}{n 5 ^{n}}, \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = \frac{n 5 ^{n}}{( n + 1 ) 5 ^{n + 1}} = \frac{n}{5 ( n + 1 )} n \to \infty \frac{1}{5}

R = 5

x = 5 : n = 1 \sum \infty \frac{1}{n} diverges x = - 5 : n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n} converges

E = [- 5, 5)

The ratio between the absolute values of two consecutive coefficients tends to $\frac{1}{5}$ , so by the ratio test the radius is the reciprocal, that is, $R = 5$ . The series converges absolutely for $∣ x ∣ < 5$ and does not converge for $∣ x ∣ > 5$ ; the two endpoints remain to be examined. At $x = 5$ the powers cancel with the factor $5^{n}$ and the harmonic series remains, which diverges. At $x = - 5$ the alternating harmonic series remains: the term $\frac{1}{n}$ is decreasing and tends to $0$ , so by the Leibniz test the series converges. The set of convergence therefore contains the left endpoint and not the right one.

Exercise 2

Determine the radius and the set of convergence of the series

n = 0 \sum \infty \frac{n + 1}{2 ^{n}} (x - 3)^{n} .

a_{n} = \frac{n + 1}{2 ^{n}}, \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = \frac{n + 2}{2 ^{n + 1}} \cdot \frac{2 ^{n}}{n + 1} = \frac{n + 2}{2 ( n + 1 )} n \to \infty \frac{1}{2}

R = 2, open interval (1, 5)

x = 5 : n = 0 \sum \infty (n + 1) diverges x = 1 : n = 0 \sum \infty (- 1)^{n} (n + 1) does not converge

E = (1, 5)

The series is centered at $x_{0} = 3$ , so the interval of convergence is centered at $3$ . The ratio between the absolute values of the coefficients tends to $\frac{1}{2}$ , therefore $R = 2$ and the open interval is $(1, 5)$ . At $x = 5$ we have $(x - 3)^{n} = 2^{n}$ , which cancels with the denominator, and the series with general term $n + 1$ remains, which tends to $+ \infty$ : the necessary condition for convergence is not satisfied, so the series diverges. At $x = 1$ we have $(x - 3)^{n} = (- 2)^{n}$ and the series with general term $(- 1)^{n} (n + 1)$ remains, which does not tend to $0$ , so the series does not converge. Neither of the two endpoints belongs to the set of convergence.

Exercise 3

Determine the radius and the set of convergence of the series

n = 1 \sum \infty \frac{n !}{n ^{n}} x^{n} .

a_{n} = \frac{n !}{n ^{n}}, \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = \frac{( n + 1 )! n ^{n}}{( n + 1 ) ^{n + 1} n !} = \frac{n ^{n}}{( n + 1 ) ^{n}} = \frac{1}{( 1 + \frac{1}{n} ) ^{n}} n \to \infty \frac{1}{e}

R = e

x = \pm e : b_{n} = \frac{n ! e ^{n}}{n ^{n}}, \frac{b _{n + 1}}{b _{n}} = \frac{e}{( 1 + \frac{1}{n} ) ^{n}} > 1 ⟹ b_{n} \geq b_{1} = e for every n

E = (- e, e)

The ratio between the coefficients simplifies to $(1 + \frac{1}{n})^{- n}$ , which tends to $\frac{1}{e}$ by the standard limit defining the number $e$ ; hence the radius is $R = e$ . At the endpoints $x = \pm e$ the absolute value of the general term is $b_{n} = \frac{n ! e ^{n}}{n ^{n}}$ . The ratio $\frac{b _{n + 1}}{b _{n}}$ equals $\frac{e}{( 1 + \frac{1}{n} ) ^{n}}$ , and it is greater than $1$ because the sequence $(1 + \frac{1}{n})^{n}$ is increasing with limit $e$ , so it always stays less than $e$ . Therefore the sequence $b_{n}$ is increasing and all its terms are greater than or equal to $b_{1} = e$ : the general term does not tend to $0$ , so the series does not converge at either of the two endpoints. The set of convergence is the open interval $(- e, e)$ .