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

For every $n$, I have a polynomial $p_n(x)=a^{(n)}_{n-1}x^{n-1}+a^{(n)}_{n-2}x^{n-2}+\dots+a^{(n)}_0$ (the $n$ in the exponent of the coefficients is merely an index).

I can show that $\lim_{n\to\infty}\sqrt[n]{a^n_{n-1}}=C$ for some constant C, and that this sequence is rising.

I can also show that we have $\lim_{n\to\infty}\sqrt[n]{p_n(x)}=A_x$ for some constant $A_x$.

My goal is to show that $\lim_{x\to\infty}(A_x/x)\ge C$.

I cannot assume the limits (of n and x) can be interchanged, although if it's easily provable I'll be glad to hear how.

The major obstacle I fail to see how to tackle is the fact that we take the $n$-th root, but the polynomial is of degree $n-1$. Were the polynomial of degree $n$ it would be much more natural as the division by $x$ would cancel out exactly with the leading coefficient of the polynomial.

Note that the claim may be incorrect (though it's unlikely) or I might be missing some assumptions (much more likely).

share|cite|improve this question
Wait, do you mean that there's an infinite sequence $ a_0, a_1, a_2, \dots $ and an infinite number of polynomials $ p_n(x) = a_{n-1}x^{n-1} + a_{n-2} x^{n-2} + \dots + a_0 $ ? (Otherwise your first limit makes no sense given you only have a finite handful of $ a_n $'s laying around, and your second limit is always 1, 0, or undefined depending on $ p(x) $'s sign.) Also, in what sense is $ A_x $ constant if it's indexed and varying with the variable $ x $? – anon Jun 9 '11 at 7:23
Yes, you are correct, this is an infinite sequence of polynomials. – Gadi A Jun 9 '11 at 7:41
What sequence is rising (meaning increasing)? What are the hypotheses on $a_k^{(n)}$ for $k\le n-2$? – Did Jun 9 '11 at 8:02
The sequence of the $n$-th root of the coefficient. Of the other coefficients I know nothing. – Gadi A Jun 9 '11 at 8:08
The degree $n-1$ vs degree $n$ stuff is not a problem at all. But: do you assume all the coefficients $a_k^{(n)}$ to be nonnegative? Otherwise the $n$th root of $p_n(x)$ might not be defined. Or did you forget an absolute value sign? And it seems the nonnegativity assumption would make the result (trivially) true... – Did Jun 9 '11 at 9:30
up vote 0 down vote accepted

Hint: fix $D<C$ and try to show that $A_x\ge Dx$ for every $x$ large enough knowing that $a_n\ge D^n$ for every $n$ large enough. In turn, $A_x\ge Dx$ would follow from $p_n(x)\ge D^nx^n$ for every $n$ large enough or even from $p_n(x)\ge D^{n-1}x^{n-1}$ for every $n$ large enough...

share|cite|improve this answer

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.