Convergence of Double Series in Real Analysis

analysiscalculusreal-analysissequences-and-series

Let $\{a_i\}_{i\in \mathbb{N}}$ be a sequence of positive real numbers, whose series converges i.e. $\sum\limits_{i\in \mathbb{N}}a_i = a < \infty.$ Does this imply the convergence of the double series $\sum\limits_{i,j \in \mathbb{N}}a_i a_j?$

If yes, how can this be proven? If not, what are some counterexamples?

Any help is appreciated. Thanks in advance.

P.S. : A double series $\sum\limits_{i,j \in \mathbb{N}}a_i a_j$ converges to A, if for all $\epsilon > 0,$ there exists $ N_0 \in \mathbb{N}$ such that \begin{eqnarray}
\left|\sum\limits_{i=1}^m\sum\limits_{j=1}^n a_i a_j -A \right| \leq \epsilon, \qquad \text{for all } m,n \geq N_0.
\end{eqnarray}

Best Answer

Let $S_n:=\sum_{k=1}^na_k$. For all $\epsilon>0$, there exists $N$ such that $$0\le a-S_N\le\epsilon.$$ We then have $$\begin{align}\forall m,n\ge N\quad0&\le a^2-\sum\limits_{i=1}^m\sum\limits_{j=1}^na_ia_j\\ &=a^2-S_mS_n\\ &\le a^2-S_N^2\\ &=(a+S_n)(a-S_N)\\ &\le2a(a-S_N)\\ &\le2a\epsilon, \end{align}$$ which proves that $\sum\limits_{i,j \in \mathbb{N}}a_i a_j$ converges to $a^2$.

Edit, thanks to @Filippo's comment: more generally, given two converging complex sequences $S_n\to a,T_n\to b$, the "doubly indexed sequence" $(S_nT_m)$ converges to $ab$, i.e. $$\forall\epsilon>0\quad\exists N\quad\forall m,n\ge N\quad|S_nT_m-ab|\le\epsilon,$$ because $$|S_nT_m-ab|=|(S_n-a)T_m+a(T_m-b)|\le|S_n-a|\sup_{k\in\Bbb N}|T_k|+|a||T_n-b|.$$

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