This is similiar to IMC 2014 https://artofproblemsolving.com/community/c7h1279375p6724144

REDACTED

The $a_i$ are integers.

shinhue wrote: This is similiar to IMC 2014 https://artofproblemsolving.com/community/c7h1279375p6724144 But here $a_{i}$ integers spice things up

Solved with L567, mueller.25, AdhityaMV, Siddharth03

The answer is "no". Clearly $a_k\ne 0, k=0,1,\dots ,$ since otherwise $P_k(x)$ would have at most $k-1$ roots. Assume on the contrary, sequence like that exists. Suppose $P_n(x)$ has $n$ distinct real roots $x_1<x_2<\dots<x_n.$ Then $P_n'(x)$ has $n-1$ distinct real roots $y_i, x_i<y_i<x_{i+1}, i=1,2,\dots,n-1.$ So, if the claim holds for a sequence $(a_n)$ it also holds for the sequence $na_n, n=0,1,\dots .$ That's why, we can assume wlog that $a_n$ takes as large values as we want, since otherwise we would transfer the claim to $na_n.$ This means there exist infinitely many $a_n$ for which $$|a_n|\ge \max_{0\le i\le k} |a_i| .$$For these $n,$ the polynomial $P_n(x)$ cannot have a real root with magnitude larger than $2,$ that is, all roots of $P_n(x)$ are in $[-2,2].$ In the same way as above, $P_n^{(k)}(x)$ has $n-k$ distinct real roots $-2<\xi_1<\xi_2<\dots<\xi_{n-k}<2$ for each $k<n.$ Note that the constant term of $P_n^{(k)}(x)$ is $a_k k!.$ We have $$|\xi_1\xi_2\cdots \xi_{n-k}|=|a_k k!|\ge k!.$$But on the other hand, since $|\xi_i|\le 2, i=1,2,\dots,n-k,$ we get $$|\xi_1\xi_2\cdots \xi_{n-k}|\le 2^{n-k}$$which implies $2^{n-k}\ge k!.$ We put $k:=n/2$ and get $$2^{n/2}\ge (n/2)!$$which is false for sufficiently large $n,$ contradiction. Remark. The point is that a sequence of real numbers like that has to have terms as close to $0$ as we want. The same proof goes. It's also redundant to require the roots of $P_n(x)$ be distinct.

Beautiful Problem! Here's a sketch with motivation: Clearly $a_k \neq 0$ for $k > 0$, else if there will be $\leq k-1$ roots. Since we're given that a random polynomial should have real roots, this causes us to think about the size of the roots, and $P'$ (because by Rolle's every root of $P'$ lies between two consecutive roots of $P$). Now the key observation is that, we can make $a_i$ grow arbitrarily larger by this (use $a_i \rightarrow ia_i$). Now this motivates the idea of "Look at a peak", so take a peak $a_n \geq \max\{a_1,a_2, \cdots a_{n-1}\}$, now onto size of roots and vieta, For these $k$ we can not have the magnitude of roots $> 2$. so by vieta's for $P^{(k)}_{n}(x)$ $2^{n-k} \geq \text{product of roots} = |a_kk!| \geq k!$ but this gives $2^{n-k} \geq k!$ which is obviously false for large $n$.(set $k = \frac{n}{2}$), contradiction!

The answer is no; FTSOC, assume yes. Then, clearly, for each $i$, $a_i \neq 0$. For each $i$, the sum of the squares of the reciprocals of the roots of $P_i (x)$ remains fixed by Vieta, so for very large $i$, the products of the reciprocals of the roots should approach $0$. However, by Vieta, it equals $\pm \tfrac{a_i}{a_0}$, which will not approach $0$ since $|a_i| \geq 1$.