Worked examples · Sheet 1

Terms, monotonicity and boundedness of sequences

Theory used: Sequence Monotonic and bounded sequences Limit of a sequence

Exercise 1

Given the sequence $a_{n} = \frac{n}{n ^{2} + 1}$ , with $n \geq 1$ , compute its first three terms and determine whether it is monotonic and whether it is bounded.

a_{1} = \frac{1}{2}, a_{2} = \frac{2}{5}, a_{3} = \frac{3}{10}

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

n \geq 1 ⟹ - n^{2} - n + 1 \leq - 1 < 0 ⟹ a_{n + 1} < a_{n}

0 < a_{n} \leq a_{1} = \frac{1}{2}

The first terms are obtained by substituting $n = 1, 2, 3$ into the general term, and they get smaller and smaller. To prove that the sequence decreases, one studies the sign of the difference $a_{n + 1} - a_{n}$ . Bringing to a common denominator, the numerator equals $n^{3} + n^{2} + n + 1 - n^{3} - 2 n^{2} - 2 n = - n^{2} - n + 1$ . For $n \geq 1$ the numerator is at most $- 1$ , while the denominator is positive, so the difference is negative and the sequence is decreasing. All the terms are positive, and since the sequence decreases the largest one is the first; therefore $0 < a_{n} \leq \frac{1}{2}$ and the sequence is bounded.

Exercise 2

Compute the first four terms of the sequence defined by recursion by $a_{0} = 3$ and $a_{n + 1} = 2 a_{n} - 1$ . Then verify that the general term $b_{n} = 2^{n + 1} + 1$ satisfies the same rule and has the same first term.

a_{1} = 2 \cdot 3 - 1 = 5, a_{2} = 2 \cdot 5 - 1 = 9, a_{3} = 2 \cdot 9 - 1 = 17

b_{0} = 2^{1} + 1 = 3

2 b_{n} - 1 = 2 (2^{n + 1} + 1) - 1 = 2^{n + 2} + 1 = b_{n + 1}

The sequence is given by recursion, so each term is obtained from the previous one by doubling it and subtracting $1$ . Starting from $a_{0} = 3$ , the rule is applied three times, giving $5$ , $9$ , $17$ . The proposed general term equals $3$ for $n = 0$ , and substituted into the rule it gives exactly the next term $b_{n + 1}$ . The two sequences therefore have the same first term and the same rule, so they coincide and $a_{n} = 2^{n + 1} + 1$ for every $n$ . As a check, $b_{3} = 2^{4} + 1 = 17 = a_{3}$ . The sequence is neither arithmetic nor geometric, because at each step one multiplies and adds.

Exercise 3

Study the monotonicity, the boundedness and the limit of the sequence $a_{n} = (- 1)^{n} \frac{3}{n}$ , with $n \geq 1$ .

a_{1} = - 3, a_{2} = \frac{3}{2}, a_{3} = - 1, a_{4} = \frac{3}{4}

∣ a_{n} ∣ = \frac{3}{n} \leq 3

∣ a_{n} - 0∣ = \frac{3}{n} < ε ⟺ n > \frac{3}{ε} ⟹ n \to + \infty lim a_{n} = 0

The terms of

a_{n} = (- 1)^{n} \frac{3}{n}

have alternating signs and approach

0

The factor $(- 1)^{n}$ makes the terms alternately negative and positive, therefore the sequence is not monotonic; for example $a_{1} < a_{2}$ but $a_{2} > a_{3}$ . The absolute value of the terms is $\frac{3}{n}$ , which does not exceed $3$ , so the sequence is bounded. For the limit, one uses the definition with $L = 0$ . The distance of $a_{n}$ from $0$ is $\frac{3}{n}$ , and this is less than any tolerance $ε$ from $n > \frac{3}{ε}$ on. Therefore the sequence converges to $0$ , even though it approaches $0$ from alternating sides.