## An informal way to find the limit of a convergent sequence defined by recurrence relations

##### It is to be noted that the answer derived from this method could be valid, yet the reasoning process is not rigorous.

##### In this text, r with a subscript of n will be denoted by r_n (the syntax of LaTex).

**Main Text**:

Given a sequence {r_n} defined by a recurrence relation:

where n is an integer.

Plot the graph of y=f(x) by letting r_(n+1) be y and letting r_n be x. (graph 1)

Also plot the graph of y=x. (graph 2)

We can observe that if {r_n} converges to µ, there will be a point of intersection between graph 1 and graph 2 with the coordinates (µ,µ).

In addition, the reverse of this observation also works. If two graphs intersect at a point (t, t), the sequence {r_n} converges to t.

Why?

Let us consider the meaning of the point of intersection of these two graphs.

Given that

the point of intersection suggests that for this specific set of values (x, y), the relation y=x is held, i.e. r_(n+1)=r_n. The sequence thus converges to µ (µ=x=y).

However, when there are multiple points of intersection between these two graphs, how do we know to which value this sequence converges, or under what conditions {r_n} will converge to a specific value?

Consider the gradient of the graph 1 (y=f(x)),

Given that the value of x is taken within an interval [å, ß], where å and ß are the x-coordinates of two points of intersection (A and B) between graph 1 and graph 2, if the gradient of the graph 1 (y=f(x)) is greater than 1, the sequence will converge to the ß, the x-coordinate of B.

This is because gradient being greater than 1 implies y>x within this interval, i.e. r_(n+1)>r_n. This means that the sequence is an increasing sequence, so the value of r_n will increase till the value of convergence.

Similarly, if the gradient is smaller than 1, the sequence will converge to the x-coordinate of A.