Worked examples · Sheet 3

Complete study of the behavior at the endpoints

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

Exercise 1

Determine the set of convergence of the series

n = 1 \sum \infty \frac{( x - 2 ) ^{n}}{n 4 ^{n}} .

a_{n} = \frac{1}{n 4 ^{n}}, \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = \frac{n}{4 ( n + 1 )} n \to \infty \frac{1}{4} ⟹ R = 4

open interval: (- 2, 6)

x = 6 : n = 1 \sum \infty \frac{4 ^{n}}{n 4 ^{n}} = n = 1 \sum \infty \frac{1}{n} diverges

x = - 2 : n = 1 \sum \infty \frac{( - 4 ) ^{n}}{n 4 ^{n}} = n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n} converges

E = [- 2, 6)

The center is $x_{0} = 2$ and the ratio between the absolute values of the coefficients tends to $\frac{1}{4}$ , so $R = 4$ and the open interval of convergence is $(2 - 4, 2 + 4) = (- 2, 6)$ . At each endpoint the value of $x$ is substituted and the resulting numerical series is studied. At $x = 6$ we have $(x - 2)^{n} = 4^{n}$ , the power cancels and the harmonic series remains, which diverges. At $x = - 2$ we have $(x - 2)^{n} = (- 4)^{n}$ and the alternating harmonic series remains, which converges by the Leibniz test, since $\frac{1}{n}$ is decreasing and tends to $0$ . The set of convergence therefore includes only the left endpoint.

Exercise 2

Determine the set of convergence of the series

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

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

open interval: (- 3, 1)

x = 1 : n = 1 \sum \infty \frac{( - 1 ) ^{n} 2 ^{n}}{n ^{2} 2 ^{n}} = n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}} converges absolutely

x = - 3 : n = 1 \sum \infty \frac{( - 1 ) ^{n} ( - 2 ) ^{n}}{n ^{2} 2 ^{n}} = n = 1 \sum \infty \frac{1}{n ^{2}} converges

E = [- 3, 1]

The factor $(- 1)^{n}$ has absolute value $1$ , so it does not affect the radius: the ratio between the absolute values of the coefficients tends to $\frac{1}{2}$ and $R = 2$ . The series is centered at $x_{0} = - 1$ , therefore the open interval is $(- 3, 1)$ . At $x = 1$ the absolute value of the general term is $\frac{1}{n ^{2}}$ , the term of a convergent generalized harmonic series, so the series converges absolutely. At $x = - 3$ the product $(- 1)^{n} (- 2)^{n}$ equals $2^{n}$ , because the two negative signs cancel; the power cancels and the series $\sum \frac{1}{n ^{2}}$ remains, which converges. Both endpoints belong to the set of convergence, so the interval is closed.

Exercise 3

Determine the set of convergence of the series

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

a_{n} = \frac{1}{n ln n}, \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = \frac{n ln n}{( n + 1 ) ln ( n + 1 )} n \to \infty 1 ⟹ R = 1

open interval: (0, 2)

x = 2 : n = 2 \sum \infty \frac{1}{n ln n} diverges (\int_{2}^{+ \infty} \frac{d t}{t ln t} = [ln (ln t)]_{2}^{+ \infty} = + \infty)

x = 0 : n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n ln n} converges

E = [0, 2)

The ratio between the coefficients tends to $1$ , since $\frac{n}{n + 1} \to 1$ and $\frac{ln n}{ln ( n + 1 )} \to 1$ , so $R = 1$ and the open interval, centered at $x_{0} = 1$ , is $(0, 2)$ . At $x = 2$ the series $\sum \frac{1}{n ln n}$ remains: comparison with the harmonic series is not conclusive, because the term is smaller than $\frac{1}{n}$ , so the integral test is applied. The function $\frac{1}{t ln t}$ is positive and decreasing for $t \geq 2$ and one of its antiderivatives is $ln (ln t)$ , which tends to $+ \infty$ : the integral diverges, therefore the series diverges. At $x = 0$ the alternating series with term $\frac{1}{n ln n}$ remains, which is decreasing and infinitesimal, and it converges by the Leibniz test. The set of convergence is $[0, 2)$ .