Does the Sequence Converge?

[evinda]

Hello! (Wave)

Let $(a_n)$ be a sequence that satisfies the relations $a_{n+1}=\frac{a}{a_n}$, where $a>1$, and $a_0<1$. I want to show that the sequence does not converge.

I have thought to suppose that the sequence does converge. Then $a_n \to l<+\infty$.
Then we get that $l=\frac{a}{l} \Rightarrow l^2=a>1$.
Also $a_1=\frac{a}{a_0}>a>1$.

But I don't see how we get a contradiction. Could you give me a hint? (Thinking)

 

[castor28]

Hello! (Wave)

Let $(a_n)$ be a sequence that satisfies the relations $a_{n+1}=\frac{a}{a_n}$, where $a>1$, and $a_0<1$. I want to show that the sequence does not converge.

I have thought to suppose that the sequence does converge. Then $a_n \to l<+\infty$.
Then we get that $l=\frac{a}{l} \Rightarrow l^2=a>1$.
Also $a_1=\frac{a}{a_0}>a>1$.

But I don't see how we get a contradiction. Could you give me a hint? (Thinking)

We have:
$$ \begin{aligned}
a_{n+1} &= \frac{a}{a_n}\\
a_{n+2} &= \frac{a}{a_{n+1}} = a_n
\end{aligned}$$

and the sequence looks like ${x,y,x,y,\dots}$.

 

[evinda]

We have:
$$ \begin{aligned}
a_{n+1} &= \frac{a}{a_n}\\
a_{n+2} &= \frac{a}{a_{n+1}} = a_n
\end{aligned}$$

and the sequence looks like ${x,y,x,y,\dots}$.

Ah I see... But how do we justify it formally?

Do we suppose that $a_n \to l<+\infty$ ?

If so, then we have that $a_{n+1} \to l$ and $a_{n+2} \to l$.
The second relation holds. The first relation gives us that $\frac{a}{a_n} \to l$ and this implies that $\frac{a}{l}=l \Rightarrow a=l^2$. But this holds, or not? (Thinking)

How else can we justify it?

 

[HallsofIvy]

A sequence of the form x, y, x, y, ... has two subsequences, one that converges to x and one that converges to y. The sequence itself converges if and only if x= y. Here, You have [math]a_1= \frac{a}{a_0}[/math], [math]a_2= \frac{a}{a_1}= \frac{a}{1}\frac{a_0}{a}= a_0[/math], [math]a_3= \frac{a}{a_2}= \frac{a}{1}\frac{1}{a_0}= \frac{a}{a_0}[/math], etc. That is, [math]x= a_0[/math] and [math]y= \frac{a}{a_0}[/math]. This sequence converges if and only if [math]a_0= \frac{a}{a_0}[/math] or [math]a= a_0^2[/math]. Since we are told that [math]a< 0[/math] that is a contradiction.

 

[castor28]

Of course, we may also cheat a little. The question does not say that $a_0\ge 0$, only that $a_0<1$. If we take $a_0=-\sqrt{a}$, the sequence is constant and therefore convergent.:)

 

[I like Serena]

Of course, we may also cheat a little. The question does not say that $a_0\ge 0$, only that $a_0<1$. If we take $a_0=-\sqrt{a}$, the sequence is constant and therefore convergent.:)

Cheating or not, doesn't it mathematically just mean that the statement is false?
And that you have given a counter example?

 

[castor28]

Cheating or not, doesn't it mathematically just mean that the statement is false?
And that you have given a counter example?

Well, we can say that the sequence does not always converge. If this was an exam question, I would answer that the sequence converges if and only if $a_0=-\sqrt{a}$.

 

[evinda]

A sequence of the form x, y, x, y, ... has two subsequences, one that converges to x and one that converges to y. The sequence itself converges if and only if x= y. Here, You have [math]a_1= \frac{a}{a_0}[/math], [math]a_2= \frac{a}{a_1}= \frac{a}{1}\frac{a_0}{a}= a_0[/math], [math]a_3= \frac{a}{a_2}= \frac{a}{1}\frac{1}{a_0}= \frac{a}{a_0}[/math], etc. That is, [math]x= a_0[/math] and [math]y= \frac{a}{a_0}[/math]. This sequence converges if and only if [math]a_0= \frac{a}{a_0}[/math] or [math]a= a_0^2[/math]. Since we are told that [math]a< 0[/math] that is a contradiction.

Ah I see... Thank you! (Smile)

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Of course, we may also cheat a little. The question does not say that $a_0\ge 0$, only that $a_0<1$. If we take $a_0=-\sqrt{a}$, the sequence is constant and therefore convergent.:)

Cheating or not, doesn't it mathematically just mean that the statement is false?
And that you have given a counter example?

Well, we can say that the sequence does not always converge. If this was an exam question, I would answer that the sequence converges if and only if $a_0=-\sqrt{a}$.

So the statement is just true, when we consider that $0 \leq \alpha_0<1$, right?

 

[I like Serena]

So the statement is just true, when we consider that $0 \leq \alpha_0<1$, right?

Yep. (Nod)

 

[evinda]

Yep. (Nod)

Nice, thank you! (Happy)