Is an analogus function useful for proving sequence monotonicity?

[DivGradCurl]

Folks,

This is the solution I have for a problem in my textbook regarding sequences. I just need to know whether I have the right idea in mind.

Thank you very much!

We can use an analogus function to show that the sequence given by

$$a_{n+1} = \sqrt{2+a_n} \quad a_1 = \sqrt{2}$$

is increasing. Here it goes

$$y = \sqrt{2+x} = \left( 2+x \right) ^{\frac{1}{2}}$$

$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+x}}>0 \Longrightarrow a_{n+1}>a_n$$

 

[phoenixthoth]

You'd have to prove not that the function itself is increasing but that the nth iterate of x is less than the n+1st iterate of x. IF f is increasing then x<f(x) will to the job, I think, because then you could take f of both sides n times to get f^n(x)<f^n+1(x).

a_n=f^n(x).

 

[vsage]

Ok I gave it some thought and I really can't give you a definitive answer but I can ask you why you think that your analogous function is analogous? The question posed in the book will have a_infinity = 2 and in fact the function converges to the function y^2 - y - 2 = 0, whereas y = sqrt(2+x) goes well beyond 2. It just doesn't add up too well to me. I'll delete this if someone gives a better response (or maybe my incorrectness will compell someone to answer correctly)

 

[DivGradCurl]

Guys,

I've just found out it is better to use mathematical induction or graphical methods instead. Thanks for the help.

 

[Pyrrhus]

I've used Differential Calculus to find out when does it increase or decrease a sequence with an anologous function, and then just check for n>= 1 values. I see no problem using this method as long as you understand what you're doing.

 

[DivGradCurl]

Oh... really? That's good.

Thank you very much!