Vtyjkyuk

Define a sequence $a_n$ as follows: $a_1=1,a_2=sqrt7,a_3=sqrt7sqrt7, a_4=sqrt7sqrt7sqrt7, dots$

Determine if it's convergent and find its limit.

The sequence satisfies $a_n=sqrt7a_n-1$. If it's convergent with limit $a$, then $a^2=7a$, so either $a=0$ or $a=7$. We show that the sequence is monotonically increasing and bounded above by $7$ so its limit cannot be equal $0$, thus it is $7$.

Claim. $a_n < 7$ for all $n$.

Proof: Induction on $n$. The base is clear. Assume $a_n-1 < 7$. This is equivalent to saying that $7a_n-1 < 49$, which happens iff $sqrt 7a_n-1 < 7$. Then $a_n=sqrt7a_n-1 < 7$.

It remains to show $a_n$ is monotonically increasing.
Consider
$a_n/a_n-1=sqrt7/a_n-1$. We prove that this is greater than $1$. It will follow that $a_n > a_n-1$.

Since $a_n < 7$, $7/a_n-1> 1$, whence $a_n/a_n-1 > 1$. Thus the sequence is increasing and bounded above, so it's convergent. It's limit cannot be zero, so it must be $7$.

Is this a correct proof?

It all boils down to showing that $$0<a_n<7implies a_n<sqrt7a_n=a_n+1<7,$$ which is fairly obvious (geometric average).

Here's a totally different approach.

Let $T(a)=sqrt7a$. For $a geq 2$,
$$
T(a) geq sqrt14 > sqrt4 = 2
qquad
textand
qquad
|T^prime(a)| = fracsqrt72sqrta leq fracsqrt72sqrt2 = fracsqrt144 < fracsqrt164 = 1.
$$
Therefore, $T$ is a contraction on $[2, infty)$.

By the Banach fixed point theorem, $T$ has a fixed point in $X$.
You already proved that this fixed point has to be $a=7$ (since $a=0 notin X$).
Conclude by noting that $T(a_1)=T(1)geq2$.