Absolute convergence

A series converges absolutely when the series of absolute values converges; absolute convergence implies convergence.

A series of functions is uniformly absolutely convergent on a set if the series of their absolute values has uniform convergence there. A sufficient condition is the Weierstrass M-test: if $|f_n(x)|\le M_n$ for every $x$ and ${\sum }_n{M}_n<\infty$, then the tail is bounded uniformly by ${\sum }_{n\ge N}{M}_n\to 0$. This also proves uniform convergence of the original series by the Uniform Cauchy criterion.

Suppose a series of functions $\sum f_n$ has uniform absolute convergence on $E$ and $(g_n)$ is a uniformly bounded family of functions there, with common bound $M$. The triangle inequality givesThe right-hand side tends uniformly to zero as $q>p\to \infty$. The Uniform Cauchy criterion proves uniform convergence of $\sum f_n{g}_n$. Applying the same estimate to $\sum |f_n{g}_n|$ proves uniform absolute convergence too.

The shared bound over both $n$ and $x$ matters: separate bounds for each $g_n$ can grow too fast. For example, the constant functions $f_n(x)=2^{-n}$ and $g_n(x)=2^n$ have a summable original series but products identically one.

If $a_n\ge 0$ and $\sum a_n<\infty$, then $a_n\to 0$, so eventually $a_n^2\le a_n$ and $\sum a_n^2$ converges.

The Cauchy-Schwarz inequality gives

If $a_n\ge 0$, $\sum a_n<\infty$, and $p>1/2$, thenThe endpoint can fail: $a_n=1/(n(\log n{)}^2)$ gives the divergent series $\sum 1/(n\log n)$.