The sum $$\sum_{k=1}^\infty a_k b_k$$ converges when $a_k$ is monotonically decreasing and $$B_n=\sum_{k=1}^n b_k$$ is finite/bounded $\forall n$. This follows from summation by parts. I'm now wondering if this remains true when $a_n,b_n$ are a complex sequences where $|a_n|$ is monotonically decreasing and $|B_n|$ is bounded. Then \begin{align} \left|\sum_{k=1}^n a_k b_k\right| &= \left|a_{n+1}B_n + \sum_{k=1}^n B_k (a_k - a_{k+1}) \right| \\ &\leq \left|a_{n+1}B_n\right| + \sum_{k=1}^n \left|B_k\right| \left|a_k - a_{k+1} \right| \end{align} and the first term vanishes in the limit $n\rightarrow \infty$ as in the original proof. Again $|B_k|$ is bounded by $M$ and I'm now wondering what we can do with $|a_k-a_{k+1}|$ since they are not real monotonically decreasing the absolute value matters and can not be left out. If I split in absolute and phase presentation \begin{align} \sum_{k=1}^n |a_k - a_{k+1}| = \sum_{k=1}^n \left| |a_k| {\rm e}^{i\phi_k} - |a_{k+1}| {\rm e}^{i\phi_{k+1}} \right| \end{align} is this going to lead anywhere? Or is this generally not true?

If I assume $\phi_k$ vanishes as $k\rightarrow \infty$, then I can estimate \begin{align} \sum_{k=1}^n \left| |a_k| {\rm e}^{i\phi_k} - |a_{k+1}| {\rm e}^{i\phi_{k+1}} \right| &= \sum_{k=1}^n \left| |a_k| \left(1 + {\cal O}(\phi_k)\right) - |a_{k+1}| \left(1 + {\cal O}(\phi_{k+1})\right) \right| \\ &\leq \sum_{k=1}^n \left| |a_k| - |a_{k+1}| \right| + \left| |a_k|{\cal O}(\phi_k) - |a_{k+1}| {\cal O}(\phi_{k+1})\right| \end{align} where ${\cal O}(\phi_k)={\rm e}^{i\phi_k}-1$. Since $|a_k|$ is monotonically decreasing the first term telescopes.

If $|a_k|$ and $\phi_k$ are differentiable, the second term can be written as \begin{align} |a_k|\left({\rm e}^{i\phi_k}-1\right) - |a_{k+1}|\left({\rm e}^{i\phi_{k+1}}-1\right) &\sim - \frac{\rm d}{{\rm d} k} |a_k|\left({\rm e}^{i\phi_k}-1\right) \\ &= - \left\{ |a_k|'\left({\rm e}^{i\phi_k}-1\right) + i|a_k|\phi_k' \, {\rm e}^{i\phi_k} \right\} \\ &\sim -i \left\{ |a_k|' \phi_k + |a_k| \phi_k' \right\} \end{align} so if e.g. $|a_k| \sim k^{-\delta}$ and $\phi_k \sim k^{-\epsilon}$ as $k\rightarrow \infty$ this leads to $$|a_k|\left({\rm e}^{i\phi_k}-1\right) - |a_{k+1}|\left({\rm e}^{i\phi_{k+1}}-1\right) \sim k^{-1-\delta-\epsilon}$$ whose sum converges absolutely for $\epsilon>0$, $\delta>0$ and $n\rightarrow \infty$.

Is this correct?


With $|a_n|$ monotonically decreasing and $|B_n|$ bounded, $\sum_n a_n b_n$ may fail to converge even in the real case (take $a_n=(-1)^n/n$ and $b_n=(-1)^n$, say).

  • $\begingroup$ So one has to consider each case separately without some generalization? Or are there general classes of complex sequences where this is true? I mean in your case $\phi_k=\pi k$ which is not quite what I was thinking about (see edit). $\endgroup$ – Diger Dec 6 '18 at 22:16
  • $\begingroup$ Well, it depends. Abel's test, which is closely related, is valid for complex $a_n$ and real $b_n$ (in the presented form). But generally your setup is too broad (this has links to some curious examples). $\endgroup$ – metamorphy Dec 6 '18 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.