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2019-02-17, 15:12   #3
Dr Sardonicus

Feb 2017
Nowhere

1101100111112 Posts

Quote:
 Originally Posted by rudy235 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