Exercise 2.8.6

Question

\begin{enumerate}[label=(\alph*)] 
    \item Assuming the hypothesis - and hence the conclusion - of Theorem 2.8.1, show that $\sum^\infty_{k=2}d_k$ converges absolutely.

\item Imitate the strategy in the proof of Theorem 2.8.1 to show that $\sum^\infty_{k=2}d_k$ converges to $S = \lim_{n\to\infty}s_{nn}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Note that $\sum^n_{i=1}\sum^n_{j=1}a_{ij}$ contains all of the terms of $\sum^\infty_{k=2}d_k$, and thus by comparison to $\sum^n_{i=1}\sum^n_{j=1}|a_{ij}|$, $\sum^\infty_{k=2}|d_k|$ must converge. \item What we need to show is that for all $\epsilon>0$ there exists $N$ such that for all $n > N$, $|S - \sum^n_{k=2}d_k| < \epsilon$. Note first that $ \sum^n_{k=2}d_k $ contains all the elements of $s_{pp}$ when $p \leq n / 2$, and that $s_{qq}$ contains all the elements of $\sum^n_{k=2}d_k$ as long as $q \geq n-1$. Since $(s_{nn}) \to S$, for arbitrary $\epsilon_1 > 0$ we can choose $n_1$ large enough such that $|s_{n_1 n_1} - S| < \epsilon_1$. If we choose $N = 2n$ then whenever $n > N$, $\sum^n_{k=2}d_k$ will contain all terms in $s_{n_1}s_{n_1}$, and if we choose $m = n - 1$ then $s_{mm}$ will contain all terms in $\sum^n_{k=2}d_k$. Thus $$ \sum^n_{k=2}|d_k| - t_{n_1n_1} \leq t_{mm} - t_{n_1n_1} $$ where $t_{nn}$ was defined near the start of the proof of Theorem 2.8.1. Moreover since $t_{nn}$ converges (as proved in Exercise 2.8.3a) and is thus a Cauchy sequence, for arbitrary $\epsilon_2 > 0$, we can also choose $n_1$ large enough to ensure for any $m_1 > n_1$, $t_{m_1 m_1} - t_{n_1 n_1} < \epsilon_2$. Putting it all together, choosing $\epsilon_1 = \epsilon_2 = \epsilon/2$, $n_1$ large enough to satisfy the two conditions discussed above, and $N = 2n_1$: $$ \begin{aligned} \left|S - \sum^n_{k=2}d_k\right| &\leq |S - s_{n_1n_1}| + \left|\sum^n_{k=2}d_k - s_{n_1n_1}\right| \ &< \epsilon_1 + \left|\sum^n_{k=2}|d_k| - t_{n_1n_1}\right| \ &\leq \epsilon_1 + t_{mm} - t_{n_1n_1} < \epsilon_1 + \epsilon_2 \ &= \epsilon \end{aligned} $$ completing the proof. \end{enumerate}

Exercise 2.8.6

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Exercise 2.8.6

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