Question

How to find the limit of the sequence as \[n\to\infty\]

Answer 1

\[P_{n+1}=\frac{P^2_n-1}{3}\]

Answer 2

what is P?

Answer 3

this is the explicit form ,P is not given,thats what we should find in order to evaluate limit

Answer 4

but maybe we can find the limit without finding P

Answer 5

\(P=\dfrac{-1}{3}\)

Answer 6

lim_(x->infinity) 1/3 (-1+x^2) = infinity

Answer 7

@Jonask what is \[P _{0}\]?

Answer 8

you can give your own P_0,but in the example it is given,by you

Answer 9

it is clear that we are solving a quadratic equation
\[x^2-3x-1=0\]

since if we let \(P_{n+1}\) be
\[P_{n+1}=\sqrt{1+3\sqrt{1+3\sqrt{1+....}}}\implies P_{n+1}^2=1+3P_n\]

Answer 10

help eith my q pls

Answer 11

Yes solve x and show that the limit works using \(\epsilon -\delta \) stuff

Answer 12

We can apply that method iff we prove P_n is monotone decreases/ increases. You skip that step, your logic is not valid

Answer 13

When using induction to prove it, then you can assume that the sequence converts to somewhere, namely x, then every subsequence converts to the same point, then x^2 -3x-1 =0 is the next step to solve for limit point. If it is increasing, reject the negative value. If it is decreasing, reject the positive one. That's what I think.

Answer 14

yes.that is ideal reasoning,so in this case the solution to the quadratic problem is the limit?

Answer 15

@Jonask I think that: letting P_0=Alpha,
if \[1<\alpha <\frac{ 3+\sqrt{13} }{ 2 }\]
then for all n we have:
\[P _{n}<\alpha \]
furthermore the sequence is decreasing, so that sequence is bounded below, and it converges when n--> infinity. The limit L, can be obtained soling the subsequent quadratic equation:
\[L=\frac{ L ^{2}-1 }{ 3 }\]