Vtyjkyuk

The sequence $A_n$ is described as-
beginalign*
A_1 & = sqrt2\
A_n+1 & =(2+A_n)^0.5
endalign*
where $A_n$ is the $n$th term in the sequence.

I can prove by induction that this sequence is bounded by above by $2$ and then find the limit to be $2$. But in order to assume that the limit exists, I need to show that it is also increasing so I can use the monotone convergence theorem. It is easy to see why it is increasing, but how do I prove it?

Prove by induction that $A_n leq 2$. Then note that $(A_n-frac 1 2)^2 leq frac 9 4$. When expanded this reduces to $A_n+1 geq A_n$.

We need to prove that

$$A_n+1=sqrt2+A_nge A_n iff A_n^2-A_n-2 le 0 iff -1le A_n le2$$

then we have $A_n ge 0$ and by inducion we can prove that $A_nle 2$ indeed

• base case: $A_1=sqrt 2le 2$
  


• induction step: assuming true $A_nle 2$ we need to prove that $A_n+1le 2$ then

$$A_n+1=sqrt2+A_n lesqrt2+2=2$$

If we can assume $A_n, A_n+1 > 0$ then

$A_n+1 > A_n iff A_n+1^2 > A_n^2$

And $A_n+1^2n = ((2 + A_n)^frac 12) = 2 + A_n$

If we can assume $A_n < 2$ then $A_n+1^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.

And that's that.

.....

We do have to prove that $A_n > 0$ but that's trivial:

By induction: Base case: $sqrt 2 > 0$. Induction step. If $A_n >0$ then $A_n + 2 > 0$ so $A_n+1 > sqrtA_n + 2 > 0$.

And we have to prove $A_n < 2$ but that's easy:

By induction: Base case: $sqrt 2 <2$. Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_n+1 = sqrt2+A_n < sqrt 4 = 2$.

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If we start from the begining we can to it is one fell swoop:

Proposition: $0 < A_1 < A_2 < .... A_n < A_n+1 <.... < 2$.

Base Case: $0 < A_1 = sqrt 2 < 2$

Induction case: Suppose $0 < A_n < 2$ then

$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$

so $0< sqrtA_n^2 < sqrtA_n + 2 < sqrt4$

and $0< A_n < A_n+1 < 2$.

A proof by contradiction.

Suppose to the contrary that for some term, we have $A_k+1 leq A_k$. Suppose that this is the first instance for which this occurs. Now, we have that

$$A_k+1 = sqrt2+A_k,$$ so that by using our assumption,
$$ 2 + A_k= A_k+1^2 leq A_k^2.$$

Now, since $$A_k = sqrt2+A_k-1,$$
then also $$A_k^2 = 2+A_k-1.$$
So, plugging in above, we have that
$$2+A_k leq A_k^2 = 2+A_k-1.$$

In other words, $$A_k leq A_k-1.$$

This is a contradiction to the assumption that $A_k+1$ was the first instance for which this occurs. Thus, the sequence must be increasing.