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Can we show that there exist non-zero coefficients $a_n$ such that

$\sum\limits_{n=0}^\infty a_n k^n = 0$ for any $k \in \{0,1,2,\dots\}$ and $a_n$ independent of $k$?

I know that for power series with $k \in \mathbb{R}$ the only solution is $a_n = 0$. I tried to come up with solutions by finding $a_n$ such that $\sum\limits_{n=1}^\infty a_n k^n$ converges (such as $a_n = \frac{1}{n^n}$) and setting $a_0 = - \sum\limits_{n=1}^\infty a_n k^n$. However, in my examples, the sum $\sum\limits_{n=1}^\infty a_n k^n$ depends on $k$.

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Suppose $f(x)$ is an analytic function with a Taylor series $\displaystyle\sum_{n = 0}^{\infty}a_nx^n$.

Then, $\displaystyle\sum_{n = 0}^{\infty}a_nk^n = f(k)$ for all $k = 0,1,2,\ldots$, so you just need $f(x)$ to have zeros at $x = 0, 1, 2, \ldots$.

Can you think of an analytic function that is zero at all integers?

Try $f(x) = \sin(\pi x)$.

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Just look at the Taylor series for $\sin (\pi x)$ namely $\sum\limits_{n=0}^{\infty} (-1)^{n-1} \frac {(\pi x)^{2n-1}} {(2n-1)!}$

Edit: if you want an example where no $a_n$ is $0$ just look at the series expansion of $\sin (\pi x)+\sin (\pi x^{2})$.

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  • $\begingroup$ This has every other $a_n=0.$ I don't know if OP minds, or wants all nonzero $a_n.$... Maybe a slight alternation including a cosine would do that. $\endgroup$
    – coffeemath
    Sep 17, 2019 at 7:29
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    $\begingroup$ @coffeemath I have edited my answer. $\endgroup$ Sep 17, 2019 at 7:39

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