Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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

share|cite|improve this answer
Oh wow, I missed the obvious by about a mile. Thanks! – JasonMond Aug 21 '12 at 13:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.