Convergence of $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$ Define a sequence $a_n$ as follows: $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$
Determine if it's convergent and find its limit.
The sequence satisfies $a_n=\sqrt{7a_{n-1}}$. 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 $a_n=\sqrt{7a_{n-1}} < 7$.
It remains to show $a_n$ is monotonically increasing.
Consider 
$a_n/a_{n-1}=\sqrt{7/a_{n-1}}$. 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?

Here's a totally different approach.
Let $T(a)=\sqrt{7a}$. For $a \geq 2$,
$$
T(a) \geq \sqrt{14} > \sqrt{4} = 2
\qquad
\text{and}
\qquad
|T^{\prime}(a)| = \frac{\sqrt{7}}{2\sqrt{a}} \leq \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4} < \frac{\sqrt{16}}{4} = 1.
$$
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 $T(a_1)=T(1)\geq2$.
