Convergence of the series with a shifted sequence of terms Consider the space
$$
\ell^2 = \left\{ x=(x_n)_{n \in \mathbb{N}} : \sum_{ n \in \mathbb{N} } x_n^2 < \infty \right\}
$$
and for a strictly positive sequence $a=(a_n)_{n \in \mathbb{N}}$ consider the weighted space
$$
\ell^2_a = \left\{ x=(x_n)_{n \in \mathbb{N}} : \sum_{ n \in \mathbb{N} } a_n x_n^2 < \infty \right\}.
$$

Under which conditions is the family $\ell^2_a$ shift-invariant, or, in other words, if $x=(x_n)_{n \in \mathbb{N}} \in \ell^2_a$, when does it follow that $x'=(x_{n+1})_{n \in \mathbb{N}} \in \ell^2_a$? Does the summability of the sequence of weights $a=(a_n)_{n \in \mathbb{N}}$ affect the answer?

We can write
$$
\sum_{n \in \mathbb{N} } a_n x_{n+1}^2 = \sum_{n \in \mathbb{N}} a_{n+1} x_{n+1}^2 \frac{a_n}{a_{n+1}}.
$$
It then seems that series converges as long as
$$
\sup_{n \in \mathbb{N}} \frac{a_n}{a_{n+1}} < \infty. \tag{1}
$$
Is it possible to come up with a necessary and sufficient condition?

Addendum:
An example of a summable sequence of weights satisfying $(1)$ would be $a_n = 2^{-n}$ (or $e^{-n}$), $n \in \mathbb{N}$. An example of non-summable sequence of weights satisfying $(2)$ would be $a_n = 2^n$ (or $e^n$), $n \in \mathbb{N}$.
As suggested in the comments, let us now consider $a_n = e^{-n^2}$ for which we have
$$
\sum_{n \in \mathbb{N}} a_n < \infty.
$$
Then
$$
\frac{a_n}{a_{n+1}} = \frac{e^{n^2+2n+1}}{e^{n^2}}=e^{2n+1},
$$
which is unbounded.
Let us further consider $x=(x_n)_{n \in \mathbb{N}}$, given by $x_n = e^{n^2/2-n/2}$. Then $a_n x_n^2 = e^{-n}$, which means that $x \in \ell_a^2$. However,
$$
a_n x_{n+1}^2 = \frac{e^{n^2+n}}{e^{-n^2}} = e^n,
$$
which means that $x'=(x_{n+1})_{n \in \mathbb{N}} \notin \ell^2_a$.
 A: Your condition (1) is actually a desired one (i.e. it is necessary and sufficient), and in order to prove it, I would suggest to consider a more general question. Let $a$ and $b$ be strictly positive sequences. What are the necessary and sufficient conditions for the inclusion $l^2_a \subset l^2_b$?
Proposition. $l^2_a \subset l^2_b$ if and only if $\sup \{b_n/a_n: n \in \mathbb N\} < \infty$.
Using this proposition it is easy to answer your initial question. Indeed consider a positive sequence $a$ and let $\hat{a}$ be a sequence defined as $\hat{a}_1 = 1$ and $\hat{a}_{n+1} = a_n$ for $n \in \mathbb N$. Then, for a sequence $x$ the conditions $x \in l^2_{\hat{a}}$ and $x' \in l^2_{a}$ are equivalent. So $l^2_a$ is shift-invariant iff $l^2_{\hat{a}} \subset l^2_{a}$ and this is the case (see proposition above) iff $\sup\{a_{n}/a_{n+1}: n \in \mathbb N\} < \infty$.
Proof of the proposition.
The ''if'' is obvious. The ''only if'' part can be proved in many ways. If you are familiar with some functional analysis, the following proof is probably the cleanest. Let $\|\cdot\|_a$ denote the standard norm on $l^2_a$. If $l^2_a \subset l^2_b$, then the inclusion map is continuous (as a simple consequence of the closed graph theorem). That means, that $\|x\|_b \le C\|x\|_a$ for some constant $C > 0$ and all $x \in l^2_a$. Applying this to the basis vectors $e^m$ (defined as $e^m_n = 0$ if $m \ne n$ and $e^m_m = 1$) we obtain that $\sqrt{b_m} \le C \sqrt{a_m}$ for all $m \in \mathbb N$. It follows that $\sup \{b_n/a_n: n \in \mathbb N\} \le C^2$.
If you would like to avoid functional analysis, you can proceed as follows. Assume the contary, i.e. $\sup \{b_n/a_n: n \in \mathbb N\} = \infty$. So there exists an increasing sequence of positive integers $k_n$ such that $b_{k_n}/a_{k_n} \ge 2^n$. Consider $x$ defined as $x_{m} = 0$ if $m \ne k_n$ for any $n \in \mathbb N$ and $x_{k_n} = a_{k_n}^{-1/2} 2^{-n/2}$. It is easy to check that $x \in l^2_a$ and $x \notin l^2_b$.
