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Old 2019-02-17, 15:12   #3
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by rudy235 View Post
I have proven heuristically that the highest number of a for which the series still converges is e1/e ~ 1.44466786100977
The term an n--> ∞ . is e (2.718281828...)
Assuming the limit x satisfies

a^{x} \;=\;x

we have

x^{\frac{1}{x}}\;=\;a\text{.}

It is an easy exercise to prove that the largest value of

y = x^{\frac{1}{x}}

for positive real x (logarithmic differentiation works nicely) is

y\;=\;e^{\frac{1}{e}}\text{.}

This occurs at

x\; =\; e\text{.}

Last fiddled with by Dr Sardonicus on 2019-02-17 at 15:14 Reason: ginxif opsty
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