One can use Dirichlet's test to prove the alternating series test quite easily. I am wondering if there is a simple proof of Dirichlet's test by assuming the alternating series test holds.
An assumption made in the hypothesis of Dirichlet's test is that one of the sequences $b_n$ be bounded. hence, $-M<b_n<M$. By the Alternating series test $\sum (-1)^n M a_n$, with $a_n$ monotically decreasing to $0$, converges. Does this lead anywhere?
It might help if you included the statement of Dirichlet's test that you have in mind. I think of it in terms of one of the sequences having bounded partial sums, not being bounded itself: if $\{a_n\}$ and $\{b_n\}$ are sequences in $\mathbf R$ such that (i) $a_1 \geq a_2 \geq \dots \geq 0$ with $a_n \rightarrow 0$ and (ii) the partial sums $\sum_{n \leq N} b_n$ are bounded, then the series $\sum a_nb_n$ converges. (We can let the $b_n$'s in Dirichlet's test be complex numbers too, but passage to real and imaginary parts reduces that to the case where all $b_n$ are real. The case of complex $b_n$ is convenient since it allows the test to be used the setting of Dirichlet series or Fourier series, such as here.)
I can't prove the answer to your question is no, but I see no mechanism to suggest it should be yes: it sounds like you're asking whether a condition about sequences with bounded partial sums can be made equivalent to the use of alternating series, but boundedness of a sequence of partial sums can occur in many situations that are not related to a sequences having alternating sign changes (see the link above). Is your question just wishful thinking or do you have evidence it might be true?
