Worked examples · Sheet 3

Series by limit comparison

Theory used: Limit comparison test Harmonic series Generalized harmonic series Comparison test

Exercise 1

Study the behavior of the series

n = 1 \sum \infty sin \frac{1}{n} .

\frac{1}{n} \to 0, n \to + \infty lim \frac{sin \frac{1}{n}}{\frac{1}{n}} = 1 ⟹ sin \frac{1}{n} \sim \frac{1}{n}

n = 1 \sum \infty \frac{1}{n} diverges ⟹ n = 1 \sum \infty sin \frac{1}{n} diverges to + \infty

For $n \geq 1$ one has $0 < \frac{1}{n} \leq 1 < π$ , therefore $sin \frac{1}{n} > 0$ and limit comparison applies. Since $\frac{1}{n} \to 0$ , the standard limit of the sine with the substitution $y = \frac{1}{n}$ gives $sin \frac{1}{n} \sim \frac{1}{n}$ . The two series then have the same behavior; the harmonic series diverges, therefore the given series diverges to $+ \infty$ .

Exercise 2

Study the behavior of the series

n = 1 \sum \infty (1 - cos \frac{1}{n}) .

n \to + \infty lim \frac{1 - cos \frac{1}{n}}{\frac{1}{n ^{2}}} = \frac{1}{2} ⟹ 1 - cos \frac{1}{n} \sim \frac{1}{2 n ^{2}}

n = 1 \sum \infty \frac{1}{n ^{2}} converges ⟹ n = 1 \sum \infty (1 - cos \frac{1}{n}) converges

The general term is nonnegative, because the cosine does not exceed $1$ . Since $\frac{1}{n} \to 0$ , the standard limit $\frac{1 - cos y}{y ^{2}} \to \frac{1}{2}$ with $y = \frac{1}{n}$ gives $1 - cos \frac{1}{n} \sim \frac{1}{2 n ^{2}}$ . Limit comparison with the generalized harmonic series of exponent $α = 2$ applies, and since $α = 2 > 1$ that series converges; hence the given series also converges.

Exercise 3

Study the behavior of the series

n = 1 \sum \infty (n n - 1) .

n n - 1 = e^{\frac{l n n}{n}} - 1, \frac{ln n}{n} \to 0

n \to + \infty lim \frac{e ^{\frac{l n n}{n}} - 1}{\frac{ln n}{n}} = 1 ⟹ n n - 1 \sim \frac{ln n}{n}

\frac{ln n}{n} \geq \frac{1}{n} for n \geq 3, n = 1 \sum \infty \frac{1}{n} diverges ⟹ n = 1 \sum \infty (n n - 1) diverges to + \infty

The root is rewritten in exponential form, $n n = e^{\frac{l n n}{n}}$ , and one observes that the exponent $\frac{ln n}{n}$ tends to $0$ by the hierarchy of infinities. The standard limit $\frac{e ^{y} - 1}{y} \to 1$ with $y = \frac{ln n}{n}$ therefore gives $n n - 1 \sim \frac{ln n}{n}$ , and for $n \geq 2$ both quantities are positive.

It remains to establish the behavior of $\sum \frac{ln n}{n}$ . For $n \geq 3$ one has $ln n \geq 1$ , therefore $\frac{ln n}{n} \geq \frac{1}{n}$ ; the harmonic series diverges, hence by the comparison test $\sum \frac{ln n}{n}$ also diverges. By limit comparison the given series has the same behavior and diverges to $+ \infty$ .