Exercise 2.8.5

Question

\begin{enumerate}[label=(\alph*)] 
    \item Show that for all $m \geq N$
    $$|(r_1 + r_2 + \cdots + r_m) - S| \leq \epsilon$$
    Conclude that the iterated sum $\sum^\infty_{i=1}\sum^\infty_{j=1}a_{ij}$ converges to $S$.

\item Finish the proof by showing that the other iterated sum, $\sum^\infty_{j=1}\sum^\infty_{i=1}a_{ij}$, converges to $S$ as well. Notice that the same argument can be used once it is established that, for each fixed column $j$, the sum $\sum^\infty_{i=1}a_{ij}$ converges to some real number $c_j$.

\end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item For any given $m$, there must be some $N_3$ such that for $n > N_3$, $k \in \mathbf{N} \leq m$,
$$ \left| r_k - \sum^n_{j=1} a_{kj}\right| < \frac{\epsilon}{2m} $$

Then there must exist some $N$ such that when $m,n \geq N$,

$$
\begin{aligned}
    \left| \sum^m_{i=1} r_i - S \right|
    & \leq \left|\sum^m_{i=1}r_i - \sum^m_{i=1}\sum^n_{j=1}a_{ij}\right| + \left|\sum^m_{i=1}\sum^n_{j=1}a_{ij} - S \right|\
    &= \left|\sum^m_{i=1} \left(r_i - \sum^n_{j=1}a_{ij}\right)\right| + \left|s_{mn} - S \right|\
    & \leq \sum^m_{i=1}\left| r_k - \sum^n_{j=1} a_{kj}\right| + \left|s_{mn} - S \right| \
    &< \sum^m_{i=1} \left(\frac{\epsilon}{2m}\right) + \frac{\epsilon}{2} = \epsilon
\end{aligned}
$$

and thus $\sum^\infty_{i=1}\sum^\infty_{j=1}a_{ij}$ converges to $S$.

\item $\sum^\infty_{i=1}|a_{ij}|$ converges for any fixed $j$ by comparison with $\sum^\infty_{i=1}\sum^\infty_{k=1}|a_{ik}|$ which converges by the hypothesis, and thus $\sum^\infty_{i=1}a_{ij}$ converges to some real number $c_j$.
Then a similar argument to (a) can be used to show that there must be some $N$such that when $n \geq N$,
    $$\left|\sum^n_{j=1}c_j - S\right| \leq \epsilon$$
and thus $\sum^\infty_{j=1}\sum^\infty_{i=1}a_{ij}$ converges to $S$.
 \end{enumerate}

Exercise 2.8.5

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Exercise 2.8.5

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