Show convergence of $\sum_{j=0}^\infty \psi_{i+j}\psi_j$ with a converse of the ratio test?

Question

I want to prove the statement: $$\sum^\infty_{j = 0} \psi_j ^2<\infty \quad\Longrightarrow\quad \forall i \in \mathbb N,\ \sum^\infty_{j = 0} \psi_{i+j}\psi_j<\infty.$$ So far, I used the converse of the ratio test to prove this: $$0<\lim_{n\rightarrow\infty}\frac{\psi_{n+1}^2}{\psi_n^2} <1\quad\Longleftrightarrow \quad 0<\lim_{n\rightarrow\infty}\bigg|\frac{\psi_{n+1}}{\psi_n}\bigg| <1 \quad\Longrightarrow\quad \forall i \in \mathbb N,\ 0<\lim_{n\rightarrow\infty}\bigg|\frac{\psi_{n+i}}{\psi_n}\bigg| <1$$ And this implies $$\exists N \in \mathbb N \quad \text{such that}\quad \forall M>N,\ 0<\bigg|\frac{\psi_{M+i}}{\psi_M}\bigg| <1 $$ Therefore, $$0<\Bigg|\sum^\infty_{j = 0}\psi_{i+j}\psi_j\Bigg| = \Bigg|\sum^\infty_{j = 0}\frac{\psi_{i+j}}{\psi_j}\psi_j^2\Bigg|\\ \leq \sum^\infty_{j = 0} \bigg|\frac{\psi_{i+j}}{\psi_i}\bigg| \psi_j^2\\ <\sum^N_{j = 0}\bigg|\frac{\psi_{i+j}}{\psi_i}\bigg|\psi_j^2 +\sum^\infty_{j = N+1}\psi_j^2<\infty.$$ I think that this proof is all okay except the converse of the ratio test. Does the converse of the ratio test hold?

Answer 1

Your proof is not valid, $\lim_{n\to\infty}\frac{\psi_{n+1}^2}{\psi_n^2}$ may not exist, even if all $\psi_n$ are non-zero. A simple example is $\psi_n^2 = 1/2^n$ if $n$ is even, and $\psi_n^2 = 1/3^n$ if $n$ is odd.

But you can use the Cauchy-Schwarz inequality: $$ \sum_{j=0}^N |\psi_{i+j}\psi_j| \le \left(\sum_{j=0}^N |\psi_{i+j}|^2 \right)^{1/2}\cdot \left(\sum_{j=0}^N |\psi_j|^2\right)^{1/2} \le \sum_{j=0}^\infty|\psi_j|^2 $$ shows that $\sum_{j=0}^\infty \psi_{i+j}\psi_j$ is absolutely convergent.