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How do you construct a sequence of functions $f_n(x)$ such that

$s = \limsup_{n\rightarrow \infty} \sqrt[n]{f_n(x)}$

for all $s > 0$?

I know it's possible to this with a different sequence

$s = \limsup_{n\rightarrow \infty} (1 + \frac{x}{n})^n$

where $x = \log(s)$.

The motivation is from proofs on radius of convergence which rely on the definition of the radius

$r = \frac{1}{\limsup_{n\rightarrow \infty} \sqrt[n]{f_n(x)}}$

and out of curiosity I tried to construct a function similar to that for $e^x$ that could map to any $s = 1/r$ but couldn't.

1 Answers 1

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If $f_n(x)=x^n$, then $\sqrt[n]{f_n(x)}=x$ and you are there.

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    Oh wow, I missed the obvious by about a mile. Thanks!2012-08-21