Can You Help Prove These Mathematical Identities?

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guysensei1
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Basically, I was messing with my calculator,
and i thought of this:

sqrt(x^sqrt(x^sqrt(x^...)))=sqrt(x)^sqrt(x)^sqrt(x)^...

where they go infinitely.
I can't prove it, so I need help.

another thing is 1/(a)+1/(a^2)+1/(a^3)... = 1/(a-1)

Perhaps the proof is something really obvious I missed?
 
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I'm not sure about the first off the top of my head, but I'll help with the second:

What you have here is a geometric series: [itex]1 + r + r^{2} + \cdots + r^{n}[/itex] with [itex]r = 1/a[/itex]

Notice that you don't have the 1, so denoting the sum as [itex]S[/itex], and since [itex](1 + r + r^{2} + \cdots + r^{n})(1-r) = 1 - r^{n+1}[/itex]

[itex]S+1 = \lim_{n \rightarrow \infty} \frac{1 - r^{n+1}}{1-r}[/itex]

This reduces to:

[itex]S+1 = \frac{a}{a-1} \Rightarrow S = \frac{a}{a-1} - \frac{a-1}{a-1} = \frac{1}{a-1}[/itex]

Edit: typo
 
Actually I think the first one is deceptively simply. You know that [itex](x^{m})^{n}= x^{mn}[/itex], so think of the square roots simply as [itex]\frac{1}{2}[/itex] exponents.

For a finite example: [itex][x^{x^{1/2}}]^{1/2} = x^{(1/2)x^{1/2}} = (x^{1/2})^{x^{1/2}}[/itex]

Of course, this logic carries for an infinite number of exponents as well.
 
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