Question

Monotonic bounded principle

Answer 1

\[a_{1}=5\]
\[a_{n+1}=\sqrt{6a_{n}-6}\]

How do I show this converges and find its limit?

(I know how to do the limit part, but I am stuck on doing proof by induction to show it is bounded below by 4, and decreasing)

Answer 2

@ganeshie8

Answer 3

wait mistake

Answer 4

\[a_{n+1}=\sqrt{6a_{n}-7}\]

Answer 5

than a_1 = 5 so a_2 = ?

Answer 6

\[a_{2}=\sqrt{6a_{1}-7}=\sqrt{23}\]

Answer 7

exactly what is less than 5 yes ?

Answer 8

that is correct

Answer 9

so and than you calculi a_3 and a_4 will get every terms less than

so a_3 is less than a_2
a_4 is less than a_3 and so much

Answer 10

what mean that a_(n+1) is < a_n

yes ?

Answer 11

but how do I show that it is bounded below by 4

Answer 12

using induction

Answer 13

oops ignore that

Answer 14

why this <= ?

Answer 15

\[ a_{n+1}^2 = 6 a_n -7 < a_n^2
\]
for n>3

Answer 16

I can't say that I followed that elie

Answer 17

how did you get n>3

Answer 18

Why?
Consider the parabola
\[
g(x)=x^2-6 x+7
\]

Answer 19

Clever!

Answer 20

It is positive for x>4, x integer

Answer 21

It is 4 not 3

Answer 22

yes @eliesaab exactly - GREAT !!!

Answer 23

Alternatively, using the dumb induction method to show it is decreasing :

Assuming \(a_{k+1} \lt a_k\), you want to show \(a_{k+2}\lt a_{k+1}\)

\(\begin{align}a_{k+2}-a_{k+1}
&=\sqrt{6a_{k+1}-7}-\sqrt{6a_{k}-7} \\~\\&= \dfrac{6(a_{k+1}-a_k)}{\sqrt{6a_{k+1}-7}+\sqrt{6a_{k}-7} } \\~\\&\lt 0\end{align}\)

Answer 24

from reading my lecture notes it claims that I should know that it is bounded below by 4 via observing, and then in the induction I should let ak=4

so is the observing just substituting values in and realising it is getting closer to 4?

Answer 25

for a similar example
he has claimed that it is bounded above by 3

then later in his induction he let ak=3 by observing that it is an upper bound
which is the part that confuses me