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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?

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  • $\begingroup$ math.stackexchange.com/questions/589288/… $\endgroup$ – lab bhattacharjee Aug 9 '18 at 18:00
  • $\begingroup$ Although that question contains a complete solution, it doesn't answer whether my reasoning is correct. $\endgroup$ – user531232 Aug 9 '18 at 18:04
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    $\begingroup$ It's simpler, just note that $a_n = 7^{\frac12+\frac14+\frac18+...+2^{1-n}}$. Now just sum the geometric series and use continuity of $x \mapsto 7^x$. $\endgroup$ – Shalop Aug 9 '18 at 18:14
  • $\begingroup$ @Shalop For me it's not simpler, actually. I was aware of that solution, but the solutions that I produce on my own tend to be of more help, so I posted this to check whether my version has any mistakes. $\endgroup$ – user531232 Aug 9 '18 at 18:19
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    $\begingroup$ Yes, the proof is correct. $\endgroup$ – saulspatz Aug 9 '18 at 18:33
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It all boils down to showing that $$0<a_n<7\implies a_n<\sqrt{7a_n}=a_{n+1}<7,$$ which is fairly obvious (geometric average).

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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 $T(a_1)=T(1)\geq2$.

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