Can You Help Prove These Mathematical Identities?

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SUMMARY

The discussion centers on proving two mathematical identities involving infinite series and exponents. The first identity, sqrt(x^sqrt(x^sqrt(x^...)))=sqrt(x)^sqrt(x)^sqrt(x)^..., is simplified using the property of exponents, leading to the conclusion that it holds true for infinite exponents. The second identity, 1/(a)+1/(a^2)+1/(a^3)... = 1/(a-1), is identified as a geometric series, and the proof is derived using the formula for the sum of an infinite geometric series, confirming its validity.

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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: 1 + r + r^{2} + \cdots + r^{n} with r = 1/a

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

S+1 = \lim_{n \rightarrow \infty} \frac{1 - r^{n+1}}{1-r}

This reduces to:

S+1 = \frac{a}{a-1} \Rightarrow S = \frac{a}{a-1} - \frac{a-1}{a-1} = \frac{1}{a-1}

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

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

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