Exercise 2.8.4

Question

\begin{enumerate}[label=(\alph*)] 
    \item Let $\epsilon > 0$ be arbitrary and argue that there exists an $N_1 \in \mathbf{N}$ such that $m, n \geq N_1$ implies $B - \frac{\epsilon}{2} < t_{mn} \leq B$.
    \item Now, show that there exists an $N$ such that
    $$|s_{mn} - S| < \epsilon$$
    for all $m, n \geq N$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $t_{mn} \leq B$ follows from the fact that $B$ is an upper bound on $\{t_{mn} : m, n \in \mathbf{N}\}$.
    Lemma 1.3.8 indicates that there exists some $p, q$ such that $t_{pq} > B - \epsilon/2$, and since $p_1 \leq p_2 \text{ and } q_1 \leq q_2 \implies t_{p_1q_1} \leq t_{p_2q_2}$, we can choose $N_1 = \max\{p, q\}$.

\item By the triangle inequality, $|s_{mn} - S| \leq |s_{mn} - s_{nn}| + |s_{nn} - S|$. Letting $n' = \min\{n, m\}$ and $m' = \max\{n, m\}$,
$$
    \left|s_{mn} - s_{nn}\right| = \left| \sum^{m'}_{i=n'} \sum^n_{j=1} a_{ij} \right| \leq \sum^{m'}_{i=n'} \sum^n_{j=1} |a_{ij}| = \left|t_{mn} - t_{nn}\right| \leq \frac{\epsilon}{2}
$$
from Exercise 2.8.4a), as long as $m,n \geq N_1$. Since $S = \lim_{n\to\infty} s_{nn}$ there exists $N_2$ such that for $n \geq N_2$, $|s_{nn} - S| < \epsilon/2$; thus picking $N = \max\{N_1, N_2\}$ ensures
    $$|s_{mn} - S| < \epsilon$$
    for all $m, n \geq N$.

\end{enumerate}

Exercise 2.8.4

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Exercise 2.8.4

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