# Prove that the following sequence is convergent: u_1=sqrt(2), u_2=sqrt(2+sqrt(2)), ldots, u_n=sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))

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In order to prove that your sequence is convergent we will use the following theorem:

A sequence of real numbers is convergent if it is monotone and bounded.

1. proof of monotonicity

We will prove this by induction:

i) x_2=sqrt(2+sqrt(2))>sqrt(2)=x_1 which proves the base.

ii) Assume that for every k leq n     x_k>x_(k-1).

iii) Now we use assumption for k=n and add number 2 to each side.

2+x_n>2+x_(n-1) hence

x_(n+1)=sqrt(2+x_n)>sqrt(2+x_(n-1))=x_n which proves inductive step.

2. proof that sequence is bounded

Since sequence is increasing we only need to find upper bound (lower bound is x_1 ). We will prove that the upper bound is 2. Again we do this by induction:

i) x_1=sqrt(2)<2

ii)Assume that x_n<2.

iii) x_(n+1)=sqrt(2+x_n)<sqrt(2+2)=2 which proves step of induction.

We have now proven that the sequence is convergent.

We can also find the limit of the sequence:

We know that sequence is convergent meaning it has limit a. This means:

lim_(n rightarrow infty) x_n=lim_(n rightarrow infty) sqrt(2+x_(n-1))

a=sqrt(2+a)                                                       (1)

by squaring this equation we get

a^2-a-2=0 solutions to this equation are a_1=2 and a_2=-1. Since only a_1=2 satisfies (1) that is our limit.



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## Related Questions

edobro | Student

By  proof of monotonicity why did you add 2 at both sides,and is there any other way of proving this which is more compactible.

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