Worked examples · Sheet 2

Radius and set of convergence with the root test

Theory used: Root 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{2 n + 1}{3 n + 1})^{n} x^{n} .

a_{n} = (\frac{2 n + 1}{3 n + 1})^{n}, n ∣ a_{n} ∣ = \frac{2 n + 1}{3 n + 1} n \to \infty \frac{2}{3}

R = \frac{3}{2}

x = \pm \frac{3}{2} : ∣ a_{n} ∣ (\frac{3}{2})^{n} = (\frac{6 n + 3}{6 n + 2})^{n} = (1 + \frac{1}{6 n + 2})^{n} n \to \infty e^{\frac{1}{6}} \neq = 0

E = (- \frac{3}{2}, \frac{3}{2})

The coefficient $a_{n}$ is an $n$ -th power, so the root test is the natural choice: the $n$ -th root removes the exponent and leaves the ratio $\frac{2 n + 1}{3 n + 1}$ , which tends to $\frac{2}{3}$ . The radius is the reciprocal, that is, $R = \frac{3}{2}$ . At the endpoints the absolute value of the general term is $(1 + \frac{1}{6 n + 2})^{n}$ ; since $n ln (1 + \frac{1}{6 n + 2}) \to \frac{1}{6}$ , this absolute value tends to $e^{\frac{1}{6}}$ , which is different from $0$ . The necessary condition for convergence is not satisfied at either of the two endpoints, so the set of convergence is the open interval.

Exercise 2

Determine the radius and the set of convergence of the series

n = 1 \sum \infty \frac{x ^{n}}{( ln ( n + 1 ) ) ^{n}} .

a_{n} = \frac{1}{( ln ( n + 1 ) ) ^{n}}, n ∣ a_{n} ∣ = \frac{1}{ln ( n + 1 )} n \to \infty 0

R = + \infty, E = R

The denominator is the $n$ -th power of $ln (n + 1)$ , so the $n$ -th root of the absolute value of the coefficient is $\frac{1}{ln ( n + 1 )}$ , which tends to $0$ because the logarithm tends to $+ \infty$ . By the convention of the root test, a zero limit means infinite radius: the series converges absolutely for every real $x$ and there are no endpoints to study.

Exercise 3

Determine the radius and the set of convergence of the series

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

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

R = \frac{1}{e}

x = \pm \frac{1}{e} : ∣ a_{n} ∣ e^{- n} = exp (n^{2} ln (1 + \frac{1}{n}) - n)

n^{2} ln (1 + \frac{1}{n}) - n = n^{2} (\frac{1}{n} - \frac{1}{2 n ^{2}} + o (\frac{1}{n ^{2}})) - n = - \frac{1}{2} + o (1)

∣ a_{n} ∣ e^{- n} n \to \infty e^{- \frac{1}{2}} \neq = 0 ⟹ E = (- \frac{1}{e}, \frac{1}{e})

The $n$ -th root of $(1 + \frac{1}{n})^{n^{2}}$ is $(1 + \frac{1}{n})^{n}$ , which tends to $e$ by the standard limit, so the radius is $R = \frac{1}{e}$ . At the endpoints the absolute value of the general term is $∣ a_{n} ∣ e^{- n}$ , which is conveniently written as the exponential of $n^{2} ln (1 + \frac{1}{n}) - n$ . Substituting the expansion $ln (1 + t) = t - \frac{t ^{2}}{2} + o (t^{2})$ with $t = \frac{1}{n}$ , the terms in $n$ cancel and the exponent tends to $- \frac{1}{2}$ ; hence the absolute value of the general term tends to $e^{- \frac{1}{2}}$ , which is different from $0$ . The necessary condition for convergence is not satisfied at the endpoints, therefore the set of convergence is the open interval.