# Proving convergence of recursive sequence

## Homework Statement

A sequence is defined recursively by the equations A1 = 1, An+1 = 1/3(An + 4). Show that {An} is increasing and An < 2 for all n. Deduce that {An} is convergent and find its limit.

## The Attempt at a Solution

i've put what i've done in this image.
http://img297.imageshack.us/img297/8858/62530295kc7.png [Broken]

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for n=1 the statement is true
now suppose it's true for a certain n
then An+1 = ...<...=2
here I used the idea that an<2

Now suppose An+1>An for some n. Use: 1/3(An+4) > An

Now for n+1, An+2=1/3(An+1 + 4)=1/3(... + 4)=... > 1/3(An+4) if and only if (solve this for An and come to a trivial solution, in example, an<2)

so now it's increasing and smaller than 2, so...
For the limit, say an+1=an and solve.

HallsofIvy
Homework Helper
for n=1 the statement is true
now suppose it's true for a certain n
then An+1 = ...<...=2
here I used the idea that an<2

Now suppose An+1>An for some n. Use: 1/3(An+4) > An

Now for n+1, An+2=1/3(An+1 + 4)=1/3(... + 4)=... > 1/3(An+4) if and only if (solve this for An and come to a trivial solution, in example, an<2)

so now it's increasing and smaller than 2, so...
For the limit, say an+1=an and solve.
That last statement, "for the limit, say An+1= An and solve" is "shorthand" for what really happens and might be misunderstood (obviously, An+1 is never equal to An). If $\alpha$ is the limit (of course, you must have first shown that the limit exists), taking the limit of both sides of the equation, $A_{n+1}= (1/3)(A_n+ 4)$ to get $lim A_{n+1}= (1/3)(lim A_n+ 4)$ which gives [itex]\alpha= (1/3)(\alpha

ofcourse, Halls is right. an+1 is not ever an, but they have the same limit as n becomes really big.