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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).

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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

1 Answer 1

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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...

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