If $a_n$ must be all $0$? 
Show that if a sequence ${a_n}$ satisfies $\sum_{n=1}^{\infty}{a_nn^k}=0$, for all $k=1,2,...$, then $a_n=0$ for all $n$.

From the series I can only get $\underset{n\rightarrow \infty}{\lim}a_nn^k=0$, $\forall k
$.
I am confused that if $a_n=e^{-n}$, then $\underset{n\rightarrow \infty}{\lim}a_nn^k=0$, $\forall k
$, so maybe there really exists a sequence $a_n$ not all $a_n=0$, s.t. $\sum_{n=1}^{\infty}{a_nn^k}=0$, $k=1,2,...$.
The $\sum_{n=1}^{\infty}{a_nn^k}=0$ can be regarded as an inner product of $(a_1,...,a_n,...)$ and $(1^k,2^k,...,n^k,...)$.
I want to use the Hilbert space $l_2$ to solve the question.
As we have $(a_1,...,a_n,...)\in l^2
$, but $(1^k,2^k,...,n^k,...)\notin l^2$, so it's useless.
Thank you for sharing your mind.
 A: Please familiarize yourself with complex analysis if necessary...
$\textbf{Preleminaries:}$
For $j \in \mathbb{N}$ set $$g_{j}(z) := \frac{\sin(2 \pi z)}{z-j}.$$
Note that $g_{j}$ has a removable singularity at $z = j$ and we can set $g_{j}(j) = 2 \pi$. Thus $g_{j}$ is an entire function that has a Taylor series centered at $0$ with rapidly decaying coefficients. Set
$$f_{j}(z) = zg_{j}(z).$$
Note that we have $f_{j}(j) = 2 \pi j$ and $f_{j}(n) = 0$ if $n \in \mathbb{Z}\setminus\{j\}.$
Set the Taylor series of $f_{j}$ around $z = 0$ to be
$$f_{j}(z) = \sum_{k=1}^{\infty}c_{j,k}z^{k}.$$
Note that the coefficients $c_{j,k}$ rapidly decay to $0$ (faster than any inverse polynomial decay) as $k \rightarrow \infty$ as $f_{j}$ is an entire function.

$\textbf{Onto the main question:}$
Note that for all $j \in \mathbb{N}$ we have the following:
$$0 = \sum_{k=1}^{\infty}c_{j,k} \sum_{n=1}^{\infty} a_{n} n^{k}$$
Because of the decay properties of $(c_{j,k})_{k=1}^{\infty}$ we have
$$0 = \sum_{k=1}^{\infty}c_{j,k} \sum_{n=1}^{\infty} a_{n} n^{k} = \sum_{n=1}^{\infty}a_{n}\sum_{k=1}^{\infty}c_{j,k}n^{k} = \sum_{n=1}^{\infty}a_{n}f_{j}(n) = 2 \pi j a_{j}. $$
Thus $a_{j} = 0$ for all $j \in \mathbb{N}.$
Feel free to make the proof more rigorous or to remove complex analysis.
