Exercise 3.8

Question

Exercise 8:
    If $\sum a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum a_nb_n$ converges.

ghostofgarborg · Accepted Answer

We first note that thm 3.42 holds for $b_n$ a monotonously increasing sequence whose limit is 0
    as well, since $(-b_n)$ then fulfills the criteria of the theorem, and $\sum a_nb_n = - \sum a_n(-b_n)$.
    
    If $\sum a_n$ converges, the partial sums form a bounded sequence. If $b_n$ is monotonic 
    and bounded it converges to a number $B$, and we get that 
    $$ \sum a_n b_n = \sum a_n(b_n - B) + B \sum a_n $$
    The first sum on the right hand side converges by thm 3.42 and the observation above. The second
    sum converges because $\sum a_n$ does. Consequently the left hand side converges.

Amit Awasthi · Answer

\begin{proof}
      Let $\{A_n\} = \sum^n_{k=0} a_n$ if $n \ge 0, A_{-1} = 0$. Choose $M$ such that $M \ge A_n$ for all $n$. Given $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $|b_p - b_q| < \epsilon/2M$ for $N < p < q$ since $\{b_n\}$ monotonic and bounded implies it is Cauchy. Then apply summation by parts:
      \begin{align*}
        \left| \sum^q_{n=p} a_nb_n ight| &= \left[
        \begin{aligned}
          &\left| \sum^{q-1}_{n=p} A_n(b_n - b_{n+1}) ight|\
          &\quad+ | A_qb_q - A_{p-1}b_p|
        \end{aligned} ight]\
        &\le M\left[
        \begin{aligned}
          &\left| \sum^{q-1}_{n=p} (b_n - b_{n+1}) ight|\
          &\quad+ | b_q - b_p|
        \end{aligned}ight]\
        &= M\left(|b_p-b_q| + |b_q-b_p|ight)\
        &= 2M|b_p-b_q| < \epsilon
      \end{align*}
      Thus, $\Sigma a_nb_n$ converges.
    \end{proof}

Exercise 3.8

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Exercise 3.8

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