TITLE: sum of alternating $|f|$ and $|f|^2$
QUESTION [5 upvotes]: Does anyone know an example of a function $f$ for which the relation
$$
\sum_{n=1}^\infty (-1)^n |f(n)| < \infty \\
\Longleftrightarrow \\
\sum_{n=1}^\infty (-1)^n |f(n)|^2 < \infty
$$
is violated?
Even though the counterexamples are somewhat valid I was more thinking about a differentiable function $f(n)$, possibly even analytic.
In a similar manner: Does the following hold
$$
\sum_{n=1}^\infty (-1)^n |f(n)|^2 < \infty \\
\Longrightarrow \quad  \sum_{n=1}^\infty \frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} < \infty
$$
where $a>0$ ? The last one is interesting, because for $a=0$ this matches the assumption and for $a \rightarrow \infty$ the sum is bounded as well since $\left|\sum_{n=1}^\infty (-1)^n\right| \leq 1$.
If it is even possible to show
$$
\sum_{n=1}^\infty \frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} \sim {\cal O}\left(a^{-1-\epsilon}\right) \qquad {\rm as} \qquad a\rightarrow \infty
$$
and $\epsilon>0$ then the integral
$$
\frac{1}{\pi} \int_{-\infty}^{\infty} \sum_{n=1}^\infty \frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} \, {\rm d}a = \sum_{n=1}^\infty (-1)^n |f(n)| < \infty
$$
is well defined and reproduces the first relation.

REPLY [9 votes]: You can put all the weight of the harmonic series on the terms with one sign. For instance 
$$
f (n)=\begin {cases}1/n,&n\ \text {odd},\\ \ \ 0,&n\ \text {even}\end {cases}
$$
makes the first series divergent and the second convergent.