# Is it possible to find coefficients $a_n$ such that the sum is always zero?

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 $$k$$.

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

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 $$\sin (\pi x)+\sin (\pi x^{2})$$.

• 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. Sep 17, 2019 at 7:29
• @coffeemath I have edited my answer. Sep 17, 2019 at 7:39