Question

This a variation of some problem that was proposed earlier. Let \(an\) be a sequence of real numbers
defined as follows:
\[
a_1=3\\
a_{n+1}=\sqrt{5 a_n-6}

\]
Show that \(a_n \ge 2.5 \) for every \(n\).
Show that \(a_n\) is decreasing.
Find its limit.

Answer 1

\(\color{black}{\displaystyle a_1=3 }\)
\(\color{black}{\displaystyle a_2=\sqrt{5a_1-6}=\sqrt{9}=3 }\)

So for \(n=1\) and \(n=2\), \(a_n=3\).

Assume \(\color{black}{\displaystyle a_n=3 }\) (for all, blah blah blah)

\(\color{black}{\displaystyle a_{n+1}=\sqrt{5a_n-6} }\)
\(\color{black}{\displaystyle a_{n+1}=\sqrt{5\times 3-6}=\sqrt{9}=3 }\)

Answer 2

So you can show by induction ((in fact you don't even need to show the bases for n=2)), that every \(n\), \(a_n=3\), and therefore \(a_n>2.5\).

Answer 3

I would doubt though that \(a_n\) is decreasing, knowing that all terms are the same.
Also, having shown \(\forall n\in\mathbb{N}\)\((a_n=3)\), we have the limit of the sequence that way.

Answer 4

Sorry, \(a_1=4\)

Answer 5

Oh, that changes everything xD

Answer 6

just informally,

\(\color{black}{\displaystyle a_1=4 \ge2.5}\)
\(\color{black}{\displaystyle a_2=\sqrt{5(4)-6}=\sqrt{14}\ge3^2 \ge2.5}\)

Assume \(\color{black}{\displaystyle a_n\ge2.5}\) blah blah blah
\(\color{black}{\displaystyle a_{n+1}\ge\sqrt{5(2.5)-6}\ge\sqrt{5(2.5)-(2.5)(2.5)}\ge2.5}\)

Answer 7

(Knowing that 2.5^2=6.25>6 )

Answer 8

\(\color{black}{\displaystyle a_1=4}\)
\(\color{black}{\displaystyle a_2=\sqrt{5(4)-6}=\sqrt{14}<3.8}\)
So it is decreasing for the first two terms.

Assume \(\color{black}{\displaystyle a_{n-1}>a_{n}}\) for all integers \(1\le k\le n\).

Since \(\color{black}{\displaystyle a_{n}=\sqrt{5a_{n-1}-6}}\) (assuming terms are real and defined),
therefore \(\color{black}{\displaystyle a_{n}\ge 0}\), which allows, \(\color{black}{\displaystyle \left[a_{n-1}>a_{n}\right]~\Longleftrightarrow \quad (a_{n-1})^2>(a_{n})^2}\).

Then,
\(\color{black}{\displaystyle (a_{n-1})^2>(a_{n})^2}\)
\(\color{black}{\displaystyle (a_{n-1})^2-(a_{n})^2>0}\)
\(\color{black}{\displaystyle 5a_n-6-(5a_{n+1}-6)>0}\)
\(\color{black}{\displaystyle 5a_n-5a_{n+1}>0}\)
\(\color{black}{\displaystyle a_n>a_{n+1}}\).

Answer 9

(Also by induction) \(a_{n+1}<a_{n}\).

Answer 10

If this limit exists, (if sequence converges to some \(x\))
\(x=\sqrt{5x-6}\)
\(x^2=5x-6\)
\(x^2-5x+6=0\)
\((x-2)(x-3)=0\)
\(x=2,3\).

so, if we prove that \(a_n>2.5\) then we will have the limit, but I am out of time right now.
(Shabbat is starting)

Answer 11

I didn't realize but I already proved that \(a_n\ge 2.5\) for all \(n\).
What can do, also is to show \(a_n>3\).

\(\color{black}{\displaystyle a_1=4>3}\)
\(\color{black}{\displaystyle a_2=\sqrt{5(4)-6}=\sqrt{14}>3}\)

Assume \(\color{black}{\displaystyle a_n>3}\) ... then,
\(\color{black}{\displaystyle a_n>3}\)
\(\color{black}{\displaystyle (a_n)^2>9}\)
\(\color{black}{\displaystyle (5a_{n+1}-6)>9}\)
\(\color{black}{\displaystyle 5a_{n+1}>15}\)
\(\color{black}{\displaystyle a_{n+1}>3}\).