Is $ \{ a_n \}_{n \in \mathbb{N}} \mapsto \sum_{n \in N} a_n x^n$ a norm on the space of sequences $\mathbb{R}^\mathbb{Z}$? The set of sequences of integers real numbers $\{ a_n \}_{n \in \mathbb{N}}$ with $a_n \in \mathbb{R}$ is not a Hilbert space  but it is a Banach space vector space.  However, it can be a Hilbert space if I give it a norm such as :
$$ \{ a_n \}_{n \in \mathbb{N}} \mapsto \sum_{n \in N} |a_n|^2 $$
and require that the norm be less than infinity.  What happens if I use a different equation.
$$ \{ a_n \}_{n \in \mathbb{N}} \mapsto \sum_{n \in N} a_n x^n$$
Could this constitute a norm on the Banach space of sequence of integers (I forget the name in the textbook).  Maybe we need to say $0 < x < 1$ and / or put absolute value signs.
"Norm" on an infinite dimensional space means we have to strict to convergent subsequences (or we could accept the case $||v|| = \infty$ as an outcome.  The vector $\vec{1} = (1,1,1,\dots)$ is has infinite norm in the first case.

In terms of sequences spaces the norm I have written is $\ell^2(\mathbb{N})$.  Certainly there is an $\ell^1(\mathbb{N})$, even though it can diverge for many sequences.
$$ \{ a_n \}_{n \in \mathbb{N}} \mapsto \sum_{n \in N} |a_n| $$
Then I am asking about a weighted version of $\ell^1$.  with $0 < x < 1$.  Is that still a norm?  
$$\ell^1(\mathbb{N}) \subseteq\left\{  \{ a_n \}_{n \in \mathbb{N}} :  \sum_{n \in N} |a_n| x^n < \infty \right\} $$
I believe the left side is strictly smaller than the right side.  There could also be an analogue of $\ell^2$:
$$\ell^2(\mathbb{N}) \subseteq\left\{  \{ a_n \}_{n \in \mathbb{N}} :  \sum_{n \in N} |a_n|^2 x^n < \infty \right\} $$
 A: You cannot define a norm of such sort on the vector space of all real sequences, $\mathbb{R}^\mathbb{N}$, as your sums don't have to converge.
However, some of the constructions you propose are indeed vector subspaces of $\mathbb{R}^\mathbb{N}$.
Let's show it for any $p \in [1, +\infty\rangle$.
Fix $x \in \langle 0, 1\rangle$ and define:
$$\ell_{x,p} = \left\{(a_n)_{n\in\mathbb{N}} \in \mathbb{R}^\mathbb{N} : \sum_{n=1}^\infty|a_n|^px^n < +\infty\right\}$$
$\ell_{x,p}$ is indeed a vector subspace of $\mathbb{R}^\mathbb{N}$:
For $(a_n)_{n=1}^\infty, (b_n)_{n=1}^\infty \in \ell_{x,p}$ we have:
$$|a_n + b_n|^p \le (|a_n| + |b_n|)^p \le \big(2\max\big\{|a_n|, |b_n|\big\}\big)^p = 2^p \max\big\{|a_n|^p, |b_n|^p\big\} \le 2^p (|a_n|^p + |b_n|^p)$$
so we get:
$$\sum_{n=1}^\infty|a_n + b_n|^px^n \le \sum_{n=1}^\infty2^p(|a_n|^p + |b_n|^p)x^n = 2^p\left(\sum_{n=1}^\infty|a_n|^px^n + \sum_{n=1}^\infty|b_n|^px^n\right) < +\infty$$
We conclude $(a_n)_{n=1}^\infty + (b_n)_{n=1}^\infty \in \ell_{x,p}$.
For $\lambda \in \mathbb{R}$ and $(a_n)_{n=1}^\infty \in \ell_{x,p}$ we have:
$$\sum_{n=1}^\infty|\lambda a_n|^px^n = |\lambda|^p\sum_{n=1}^\infty|a_n|^px^n < +\infty$$
Hence $\lambda(a_n)_{n=1}^\infty \in \ell_{x,p}$.
Since $\ell_{x,p}$ is closed under addition and scalar multiplication, we conclude that $\ell_{x,p} \le \mathbb{R}^\mathbb{N}$.
Your observation that $\ell^p(\mathbb{N}) \subseteq \ell_{x,p}$ is correct. Take $(a_n)_{n=1}^\infty \in \ell^p(\mathbb{N})$. We have:
$$\sum_{n=1}^\infty |a_n|^p\underbrace{x^n}_{\le1} \le \sum_{n=1}^\infty |a_n|^p < +\infty$$
so $(a_n)_{n=1}^\infty \in \ell^{x,p}$.

Futhermore, we can turn $\ell_{x, p}$ into a normed space in the way you suggested.
Define $\|\cdot\|_{x,p} : \ell_{x, p} \to \mathbb{R}$ as:
$$\left\|(a_n)_{n=1}^\infty\right\|_{x,p} = \sqrt[p]{\sum_{n=1}^\infty |a_n|^px^n}, \text{ for } (a_n)_{n=1}^\infty \in \ell_{x,p}$$
To show that $\|\cdot\|_{x,p}$ is a norm on $\ell_{x, p}$, it is useful to notice that for $(a_n)_{n=1}^\infty \in \ell_{x,p}$ we have $\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty \in \ell^p(\mathbb{N})$ and:
$$\left\|(a_n)_{n=1}^\infty\right\|_{x,p} = \left\|\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_p$$
where $\|\cdot\|_p$ is the standard $p$-norm on $\ell^p(\mathbb{N})$.
Indeed, $\left\|\cdot\right\|_{x,p} \ge 0$.
We have $\left\|0\right\|_{x,p} = 0$. Assume $0 = \left\|(a_n)_{n=1}^\infty\right\|_{x,p} = \left\|\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_p$. It follows that $\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty = 0$ so $a_n = 0$ for all $n \in \mathbb{N}$.
For $\lambda \in \mathbb{R}$ we have:
$$\left\|\lambda (a_n)_{n=1}^\infty\right\|_{x,p} = \left\|(\lambda a_n)_{n=1}^\infty\right\|_{x,p} = \left\|\left(\lambda a_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_p = |\lambda| \left\|\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_p = |\lambda| \left\|(a_n)_{n=1}^\infty\right\|_{x,p}$$
Finally, for the triangle inequality we have:
\begin{align}
\left\|(a_n)_{n=1}^\infty + (b_n)_{n=1}^\infty\right\|_{x,p} &=\left\|(a_n + b_n)_{n=1}^\infty\right\|_{x,p}\\
&= \left\|\left((a_n + b_n)x^{\frac{n}p}\right)_{n=1}^\infty\right\|_{p}\\
&= \left\|\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty + \left(b_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_{p}\\
&\le \left\|\left(a_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_{p} + \left\|\left(b_nx^{\frac{n}p}\right)_{n=1}^\infty\right\|_{p}\\
&= \left\|(a_n)_{n=1}^\infty\right\|_{x,p} + \left\|(b_n)_{n=1}^\infty\right\|_{x,p}
\end{align}
Hence, $\|\cdot\|_{x,p}$ is a norm and $\left(\ell_{x,p}, \|\cdot\|_{x,p}\right)$ is a normed space.
