Is there any nonzero infinite sum of polynomials that converges to zero?

Question

Is there any infinite series of polynomials of the form $\sum_{n=0}^\infty a_n x^n$ that converges to zero on the interval $[0,1]$ where at least some $a_i \neq 0$?

Answer 1

If you are asking about a power series $P(x)=\sum_{n=0}^{\infty}a_nx^n=\lim_{m\to \infty}\sum_{n=0}^ma_nx^n$ such that $P(x)=0$ for all $x\in [0,1],$ but with $a_n\ne 0$ for some $n,$ the the answer is NO.

First observe that $\infty>M=\sup_{n\ge 0}|a_n|.$ Otherwise $P(1)$ does not exist in $\Bbb R.$

Now suppose $n_0$ is the least $n\ge 0$ such that $a_n\ne 0.$ Then for any $m>n_0 $ and any $x\in (0,1)$ we have $$|\sum_{n=0}^m a_nx^n|=|a_{n_0}x^{n_0}+\sum_{n=n_0+1}^ma_nx^n|\ge$$ $$\ge |a_n|x^{n_0}-\sum_{n=n_0+1}^m|a_n|x^n\ge$$ $$\ge |a_{n_0}|x^{n_0}-\sum_{n=n_0+1}^mMx^n\ge$$ $$\ge |a_{n_0}|x^{n_0}-\sum_{n=n_0+1}^{\infty}Mx^n=$$ $$=|a_{n_0}|x^{n_0}-\frac {Mx^{n_0+1}}{1-x}=$$ $$=x^{n_0}(|a_{n_0}|-\frac {Mx}{1-x}).$$ So for any $x\in (0,1)$ that's small enough that $\frac {Mx}{1-x}\le|a_{n_0}|/2$ we have $|P(x)|\ge |a_{n_0}|x^{n_0}/2>0.$

In general: Let $A<B$ and $C=\max (|A|,|B|).$ If $P(C)$ exists and $P(x)=0$ for infinitely many $x\in [A,B],$ where $P(x) =\sum_{n=0}^{\infty}a_nx^n,$ then every $a_n=0.$

Answer 2

You do not say anything about the meaning of convergence (pointwise, uniform, $\dots$ ?).

Be that as it may, the answer is "yes". Let $p_n(x) = (1/n) x$. Then $p_n(x) \to 0$ uniformly.

Answer 3

If your question is whether there exist $a_n$'s not all $0$ such that $\sum a_nx^{n}=0$ fro all $x \in [0,1]$ the answer is no. The radius of convergence of the series must be at least $1$ and its sum is an analytic function on the open unit disk. The zeros of this function have a limit point so its identically $0$ on the unit disk. This implies that all the coefficients are $0$.