A sequence which converges for an infinity of $x$'s

Question

Consider the real sequences $(a_n) _{n\ge 1}$,$(b_n) _{n\ge 1}$,$(c_n) _{n\ge 1}$ and $p_n(x) =(x-a_n) (x-b_n) (x-c_n)$. Prove that if $p_n(x) $ converges for an infinity of values of $x$, then it converges $\forall x\in \mathbb{R} $.
My idea was to consider the set $M=\{x\in \mathbb{R}| \lim\limits_{n\to \infty} p_n(x) \in \mathbb{R} \} $. Obviously, there exist $a=\sup M$ and $=\inf M$. From here, I think that I should either prove that $a=\infty$ and $b=-\infty$ or that $p_n(x) $ is bounded and then use this somehow. Anyway, I can't make any progress on this.

Answer 1

The goal is to show that the coefficients of the polynomials converge (not necessarily the actual sequences $a_n$, $b_n$, $c_n$ though). This is also suggested above in the comments by @Helmut.

All we need is to have three distinct points where $p_n$ converges.

Expand the product in the definition of $p_n$ and write $$ \tag{P} p_n(x) = x^3 - (a_n + b_n + c_n) x^2 + (a_n b_n + a_n c_n + b_n c_n ) x - a_n b_n c_n. $$ Denote $a_n + b_n + c_n := A_n$ and $a_n b_n + a_n c_n + b_n c_n := B_n$. Now, let $x_1, x_2, x_3$ be all pairwise distinct points where $p_n$ converges. In particular, the sequence $$ p_n(x_1) - p_n(x_2) = x_1^3 - x_2^3 - A_n (x_1^2 - x_2^2) + B_n(x_1 - x_2) = \\(x_1 - x_2)[ x_1^2 + x_1x_2 + x_2^2 - A_n (x_1 + x_2) + B_n] $$ converges as a difference of two converging sequences. Since $x_1 \neq x_2$ we get that $$ \tag{1} x_1^2 + x_1x_2 + x_2^2 - A_n (x_1 + x_2) + B_n $$ converges. Notice that in $(1)$ $x_1 $ and $x_2$ can be replaced by $x_2$ and $x_3$, hence we also get that $$ \tag{2} x_2^2 + x_2x_3 + x_2^2 - A_n (x_2 + x_3) + B_n $$ converges as well. Taking the difference of $(1)$ and $(2)$ and dropping the constant (with respect to $n$) expression involving only $x_1,x_2,x_3$ we get that $$ A_n (x_2 + x_3 - x_1 - x_2) = A_n(x_3 - x_1) $$ converges. Since $x_1\neq x_3$ it follows from the last equality that $A_n $ converges, hence getting back to $(1)$ it follows that $B_n$ is also convergent. Finally, getting back to $(P)$ - the original equation of $p_n$, and using that it converges at $x_1$, along with convergence of $A_n$ and $B_n$ we get that $a_nb_nc_n$ is also convergent.

We have proved that all coefficients in $(P)$ converge, hence $p_n(x)$ converges for any $x\in \mathbb{R}$.