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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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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