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This is a cross-posting of a MSE question (which did not receive any feedback there so far).

Let $\varepsilon >0$, with $\varepsilon \neq 1$. Consider the sequence $u_n$ defined by

$$ u_n=\sum_{k=0}^n \binom{n}{k}^2 (-\varepsilon)^k $$

Is is true that $(u_n)$ takes on negative and positive values infinitely often, while $|u_n| \to +\infty$ when $n\to \infty$ ?

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  • $\begingroup$ How did this arise? $\endgroup$ – Douglas Zare Mar 28 '13 at 11:33
  • $\begingroup$ @Douglas Zare the answer will probably disappoint you : I encountered those questions when trying to prove that the above polynomial has only real roots, as a toy example of the fact that if $\sum_{k}a_kx^k$ has only real roots then so does $\sum_{k}\binom{n}{k}a_kx^k$ (see this stackoverflow question). $\endgroup$ – Ewan Delanoy Mar 28 '13 at 15:15
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This polynomial is related to a Legendre polynomial. $P_n(x) = 2^{-n} \sum {n \choose k}^2 (1-x)^{n-k}(1+x)^k$ so your sum $u_n(\varepsilon) = (1+\varepsilon)^n P_n(\frac{1-\varepsilon}{1+\varepsilon})$.

Note that for $\varepsilon \in \mathbb R^+$, $\frac{1-\varepsilon}{1+\varepsilon} \in (-1,1)$, the natural domain for the Legendre polynomials since they are orthogonal on $(-1,1)$ and have all roots in $(-1,1)$.

The zeros of Legendre polynomials are roughly evenly spaced in the following sense: Consider $P_n(\cos \theta)$ for $\theta \in [0,\pi]$. Order the roots $0 \lt \theta_1 \lt \theta_2 \lt ... \lt \theta_n \lt \pi$. Then for $k \lt n/2$,

$$ \frac{k-1/2}{n} \pi \lt \theta_k \lt \frac{k}{n+1} \pi \lt \frac{k}{n} \pi.$$

For larger $k$, the roots are symmetric about $\pi/2$ since $\theta_k = \pi - \theta_{n+1-k}.$ (Markoff and Stieltjes, see Szegö). These constraints on the roots let us be sure of the sign of $P_n(\cos (\theta))$ when $\theta \in (0,\frac{\pi}{2n}), (2\frac{\pi}{2n},3\frac{\pi}{2n}), ...\lt \pi/2$ since there are no roots in these bands, and there is one root each in $(\frac{\pi}{2n},2\frac{\pi}{2n}),(3\frac{\pi}{2n},4\frac{\pi}{2n}),...$

I believe these bands where the sign is known sweep out the whole domain slowly enough that you can pick out an alternating sequence containing any particular value. I'll fill in the details later if no one has a cleaner proof. [Done below.]

It is not true that $|u_n(\varepsilon)| \to \infty$ for all $\varepsilon$. Although the $(1+\varepsilon)^n$ factor grows exponentially rapidly, the collected zeros of the $P_n(\frac{1-\varepsilon}{1+\varepsilon})$ factors are dense, and $P_n(\frac{1-\varepsilon}{1+\varepsilon})$ is continuous, so the product is small in an interval around each zero. Given one interval where $u_{n_0}$ is small, you can find some $n_1$ and a nested interval where $u_{n_1}$ is small, etc. The intersection constructs a value where $|u_n|$ is small infinitely often.


Here is a proof that $u_n(\varepsilon)$ changes sign infinitely often when $\varepsilon \lt 1$.

Let $\alpha = \arccos \frac{1-\varepsilon}{1+\varepsilon} \lt \frac{\pi}{2}.$ Pick any $k \in \mathbb N$. If we can find $m$ so that $4k \frac{\pi}{2m} \le \alpha \le (4k+1) \frac{\pi}{2m}$ then there are $2k$ (an even number) of roots of $u_m$ smaller than $\varepsilon$ and $u_m(0) = 1$ so $u_m(\varepsilon) \gt 0$. Solving for $m$: $4k \frac{\pi}{2} \alpha^{-1} \le m \le (4k+1) \frac{\pi}{2}\alpha^{-1} $. The length of that interval constraint on $m$ is $(4k+1)\frac{\pi}{2}\alpha^{-1} - 4k \frac{\pi}{2}\alpha^{-1} = \frac{\pi}{2} \alpha^{-1} \gt 1$ so there is at least one integer between $4k \frac{\pi}{2} \alpha^{-1} $ and $(4k+1)\frac{\pi}{2}\alpha^{-1} $. Call than $m$.

Similarly, for any $k$ we can find $m$ so that $(4k+2)\frac{\pi}{2m} \le \alpha \le (4k+3)\frac{\pi}{2m}$ so that the number of roots greater than $\varepsilon$ is $2m+1$, odd, and $u_m(\varepsilon)$ is negative.

So, for $\varepsilon \lt 1, u_n(\varepsilon)$ is positive infinitely often, and negative infinitely often. This is a little more awkward when $\varepsilon \gt 1$ since the parity of the number of roots smaller than $\varepsilon$ doesn't determine the sign.


Finally, the sign alternates infinitely often for $\varepsilon \gt 1$. This corresponds to $\frac{1-\varepsilon}{1+\varepsilon} \in (-1,0)$.

Let $\alpha = \pi - \arccos \frac{1-\varepsilon}{1+\varepsilon}$. We can determine the sign of $u_m(\varepsilon)$ if $\alpha \in [(4k)\frac{\pi}{2m},(4k+1)\frac{\pi}{2m}]$ since then there are $2k$ roots of $u_m$ larger than $\varepsilon$. If $m$ is even, then $\lim_{x\to \infty} u_m(x) = \infty$ so $u_m(\varepsilon) \gt 0$ and if $m$ is odd, then $\lim_{x\to \infty} u_m(x) = -\infty$ so $u_m(\varepsilon) \lt 0$. If we can find two adjacent integers $m, m+1$ so that both are between $4k \frac{\pi}{2} \alpha ^{-1}$ and $(4k+1) \frac{\pi}{2} \alpha^{-1}$ then $u_m(\varepsilon)$ and $u_{m+1}(\varepsilon)$ have opposite signs. Let $\beta = \frac{\pi}{2 \alpha} \gt 1$. The condition on $m$ and $k$ we want is $0 \le \frac{m}{4k} - \beta \le \frac{\beta -1}{4k}$.

From the theory of Diophantine approximations, we know every real number $\beta$ has infinitely many rational approximations from above $\frac{p}{q}$ so that $0\le \frac{p}{q} - \beta \le \frac{1}{q^2}$, where we do not require $\frac{p}{q}$ to be reduced. For any such rational approximation with $q \gt \frac{4}{\beta - 1}$ hence $\frac{1}{q} \lt \frac{\beta -1}{4}$, we have $\frac{4p}{4q} - \beta \lt \frac{1}{q^2} \lt \frac{\beta-1}{4q}$. Set $m=4p, k=q$, and $u_{4p}(\varepsilon)$ and $u_{4p+1}(\varepsilon)$ have opposite signs. From the infinitely many good rational approximations from above with large enough denominators, we find infinitely many sign changes of $u_n(\varepsilon)$.

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  • $\begingroup$ Thanks for this answer! I’m certainly looking forward to your more detailed explanation. $\endgroup$ – Ewan Delanoy Mar 28 '13 at 15:17

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