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.

  • $\begingroup$ I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it $\endgroup$ – mathworker21 Dec 1 '18 at 12:09
  • $\begingroup$ True, I was thinking about a counterexample and ended the sentence wrong. $\endgroup$ – Diger Dec 1 '18 at 12:11
  • $\begingroup$ See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite. $\endgroup$ – zwim Dec 1 '18 at 13:26
  • $\begingroup$ Good example for the other way around, but unfortunately still not analytic. $\endgroup$ – Diger Dec 1 '18 at 13:34

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.


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