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.
