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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.

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1 Answer 1

up vote 3 down vote accepted

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! –  JasonMond Aug 21 '12 at 13:56

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