My little brother started fiddling around with his calculator, and noticed something curious: $$ \Large \sqrt{a\cdot\sqrt{a\cdot\sqrt{a\cdot\sqrt{a \cdot \ldots}}}} = a $$ So I went ahead and wrote a proof for it, which follows:
Let us consider the expression: $$ \large \sqrt{a\cdot\sqrt{a\cdot\sqrt{a\cdot\sqrt{a \cdot \ldots}}}} $$ If we rewrite the expression as $$ \Large a^{\frac{1}{2} log_{a} \left( a\cdot\sqrt{a\cdot\sqrt{a\cdot\sqrt{a \cdot \ldots}}} \right)} $$ we can deduce that: $$ \frac{1}{2} log_{a} \left( a\cdot\sqrt{a\cdot\sqrt{a\cdot\sqrt{a \cdot \ldots}}} \right) = \frac{1}{2} \left( log_{a}(a) + \frac{1}{2} \left( log_{a}(a) + \frac{1}{2} \left( log_{a}(a) + \ldots \right) \right) \right) $$ Then, \begin{align} \frac{1}{2} \left( log_{a}(a) + \frac{1}{2} \left( log_{a}(a) + \frac{1}{2} \left( log_{a}(a) + \ldots \right) \right) \right) &= \frac{1}{2}log_{a}(a) + \frac{1}{4}log_{a}(a) + \frac{1}{8}log_{a}(a) + \ldots \\ \frac{1}{2}log_{a}(a) + \frac{1}{4}log_{a}(a) + \frac{1}{8}log_{a}(a) + \ldots &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = \sum_{n=1}^{\infty} \frac{1}{2^{n}} \end{align} And therefore: $$ \Large a^{\frac{1}{2} log_{a} \left( a\cdot\sqrt{a\cdot\sqrt{a\cdot\sqrt{a \ldots}}} \right)} = a^{\sum_{n=1}^{\infty} \frac{1}{2^{n}}} $$ But since $$ \large \sum_{n=1}^{\infty} \frac{1}{2^{n}} $$ is a convergent geometrical series, then we can say $$ \sum_{n=1}^{\infty} \frac{1}{2^{n}} = \frac{1}{2-1} = 1 $$ And finally: $$ \Large \sqrt{a\cdot\sqrt{a\cdot\sqrt{a\cdot\sqrt{a \cdot \ldots}}}} = a^{\sum_{n=1}^{\infty} \frac{1}{2^{n}}} = a $$ For all $a>0$
My question is, can I consider this proof valid? Did I overlook/assume something? I don't want to give my brother a broken proof for what he found.
I also wrote a generalization to this, but i won't write it in this question.
