## - Types of Sequences

A sequence is a list of numbers which obey a particular pattern. Each number in the sequence is called a term of the sequence. These are usually denoted $$u_1,u_2,u_3,…,u_{n-1}, u_n$$ where $$u_1$$ is the first term, $$u_2$$ is the second term and $$u_n$$ is the $$n^{th}$$ term. In some cases the sequence can be defined by a formula – an expression for the $$n^{th}$$ term.

A sequence can be classified as being convergent, divergent, oscillating or periodic.

CONVERGENT SEQUENCES

Convergent sequences as the name suggests converge to a definite limit. Convergent sequences as the name suggests converge to a definite limit.

$$\displaystyle \lim_{n\to\infty}u_n=l$$

The sequence above is convergent because it is tending to a value.

The sequence above is OSCILLATING and converges.

DIVERGENT SEQUENCES

Divergent sequences are sequences which are not convergent.
The sequence above diverges since it does not converge to any specific value.
This oscillating sequence above is divergent.
This divergent sequence is PERIODIC as it consists of a set of values which are constantly repeated. The repeating pattern of the sequence consists of three values therefore the sequence is said to have a period of 3.

CONVERGENCE OF A SEQUENCE
LESSON 1
Determine which of the following functions is convergent or divergent. If the sequence is convergent, determine the limit of the sequence.
(a) $$\displaystyle u_n={3n\over n+1}$$
(b) $$\displaystyle u_n={n^3\over n^4-7}$$
(c) $$\displaystyle u_n={1-2n\over \sqrt{3n+4}}$$
(d) $$\displaystyle u_n={\ln ⁡n\over n^3}$$
SOLUTION
If there is $$l$$ such that $$\displaystyle \lim_{n→\infty} u_n=l$$ then $$u_n$$ is convergent.
(a) $$\displaystyle \lim_{n→\infty} u_n=\lim_{n→\infty}⁡ {3n\over n+1}$$
$$\displaystyle =\lim_{n→\infty}⁡ {{3n\over n}\over {n\over n}+{1\over n}}$$
$$\displaystyle =\lim_{n→\infty}⁡ {3\over 1+{1\over n}}$$
$$=3$$
$$u_n$$ converges and it converges to 3

(b) $$\displaystyle \lim_{n→\infty} u_n=\lim_{n→\infty} {n^3\over n^4-7}$$
$$\displaystyle =\lim_{n→\infty}⁡ {{n^3\over n^4}\over {n^4\over n^4} -{7\over n^4}}$$
$$\displaystyle =\lim_{n→\infty} {{1\over n} \over 1-{7\over n^4}}$$
$$=0$$
$$u_n$$ is convergent and it converges to 0.

(c) $$\displaystyle \lim_{n→\infty}⁡ {1-2n\over \sqrt{3n+4}}=\lim_{n→\infty}{\sqrt{(1-2n)^2}\over \sqrt{3n+4}}$$
$$\displaystyle =\lim_{n→\infty} ⁡\sqrt{{4n^2-4n+1\over 3n+4}}$$
$$\displaystyle =\lim_{n→∞}⁡ \left({4n^2-4n+1\over 3n+4}\right)^{{1\over2}}$$
$$\displaystyle=\lim_{n→∞}⁡ {{4n^2\over n^2} -{4n\over n^2} +{1\over n^2}\over {3n\over n^2}+{4\over n^2}}$$
$$\displaystyle =\lim_{n→∞}⁡{4-{4\over n}+{1\over n^2} \over {3\over n}+{4\over n^2}}$$
DOES NOT EXIST
Not convergent

(d) $$\displaystyle \lim_{n→∞}⁡ {\ln ⁡n\over n^3}=\lim_{n→∞}{⁡{1\over n}\over 3n^2}$$By L’Hopitals
$$\displaystyle =\lim_{n→∞}⁡ {1\over 3n^3}$$
$$=0$$
Convergent and converges to 0.