# Dirichlet test for complex sequences

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).
• 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). – Diger Dec 6 '18 at 22:16
• 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). – metamorphy Dec 6 '18 at 22:50