I am working in some topic and got the following problem.
We denote $$l^2 \left( \mathbb{Z^+}\right)= \left \{ x = \left(x_0, x_1, x_2, x_3, \dots \right) st. \left \| x\right \|^2 = \sum_{n=0}^{\infty} |x_n|^2 < \infty\right \}$$.
Let $\alpha$ and $\beta$ be two positive numbers such that $\alpha < \beta$.
We define $\left \{ \alpha_n \right \} = \left \{ \alpha, \beta, \alpha, \alpha, \beta, \beta, \alpha, \alpha, \alpha, \beta, \beta, \beta, \alpha, \alpha, \alpha, \alpha, \beta, \beta, \beta, \beta, \dots\right \}$.
Taking vector $x$ as the folowing
\begin{align} x_0 &= 0\\ x_1 &= x_1\\ x_{n+1} &= \frac{\prod^{n}_{j=1}\alpha_j x_1}{\lambda^n} \quad \text{for some} \quad \lambda \in \mathbb{C}. \end{align}
More precisely,
$x= \left ( 0, 1, 1, \frac{\alpha}{\beta}, \frac{\alpha^2}{\beta^2}, \frac{\alpha^2}{\beta^2},\frac{\alpha^2}{\beta^2},\frac{\alpha^3}{\beta^3},\frac{\alpha^4}{\beta^4},\frac{\alpha^5}{\beta^5},\frac{\alpha^5}{\beta^5},\frac{\alpha^5}{\beta^5},\frac{\alpha^5}{\beta^5},\frac{\alpha^6}{\beta^6},\frac{\alpha^7}{\beta^7},\frac{\alpha^8}{\beta^8}, \frac{\alpha^9}{\beta^9},\frac{\alpha^9}{\beta^9},\dots\right )$.
By some simple calculation, we see that $\|x\| < \infty$ for $|\lambda| > \beta$
However, if $\lambda = \beta$, does $x \in l^2 \left( \mathbb{Z^+}\right)$?
I got trouble in that question.
Thank you for your time.
The sequence is in $\ell^2$, and in fact it's even in $\ell^1$. Let $r=\alpha/\beta \in (0, 1)$, so the series looks like $$ r + r^2+r^2 + r^3 + r^4+r^5+r^5+r^5+r^5+r^6+r^7 + \cdots $$ Let's say that the term $r^k$ appears $m_k$ times here. Since the terms are nonnegative, we only need a uniform bound on some infinite sequence of partial sums. The partial sums $$\sum_{k=1}^T m_k r^k$$ are particularly convenient. Since $r<1$, these are uniformly bounded if $\sum_{k=1}^\infty m_kr^k$ converges, which is guaranteed if $m_k$ is bounded by some power of $k$. Here $m_k\le k$, so we are done.
Another way to explain this is: for a series of nonnegative terms, we can group them any way we want without changing the convergence/divergence status of the series. Here it helps to group all equal terms together.