If the absolute value of an infinite series is bounded above, then the infinite series absolutely converges?

Question

I am attempting to follow this simple example found in Further Linear Algebra by Blyth & Robertson page 14. It deduces that a sequence of partial sums must be absolutely convergent when bounded above. But I don't believe that the partial sums are necessarily monotonically increasing (there could be negative summands within each partial sum). I'm missing something obvious.

screenshot from book example

In case the image does not appear, here is the context: The book is using Cauchy Schwarz to show that the standard inner product is well-defined (convergent) in the space of square summable real sequences.

Let $a = (a_1, a_2, ... )$ and $b = (b_1, b_2, ...)$ be such that $\sum_{i=1}^{\infty} |a_i|^2$ and $\sum_{i=1}^{\infty} |b_i|^2$ converge.

Define the following sequences of partial sums:

$p_n = \sum_{i=1}^{n}a_i b_i$ and $q_n = \sqrt{\sum_{i=1}^{n} |a_i|^2 \sum_{i=1}^{n} |b_i|^2 }$ for $n=1, 2, ...$

$(q_n)$ converges because $a$ and $b$ are square summable. By Cauchy Schwarz, $|p_n| \leq q_n$. The book then deduces that $(p_n)$ is therefore an absolutely convergent sequence. I don't see how this could be applying the bounded monotone convergence theorem because the sum $p_n$ could contain negative summands, making the sequence $(|p_1|, |p_2|, ...)$ not necessarily monotonic.

Answer 1

It is certainly not true that an upper bound on the absolute value of partial sums implies convergence of a series (consider $1-1+1-1\dots$).

Therefore, I think from context we can infer that what the author intended (or should have intended) to argue was as follows:

Define $r_n=\sum_{i=1}^n |a_ib_i|$, and show that $r_n\leq q_n$. This follows from Cauchy Schwarz just as it did for $|p_n|$.

Therefore, since $r_n$ (which is monotone, and bounded by $\lim_{n\to\infty} q_n$) converges, $p_n$ converges as well, by the result that absolute convergence of a series implies convergence (applied here to $\mathbb R$).