I'm trying to learn the convergence of $\sum_{n\ge 1} z^n/n$ when $|z|=1$ but $z\not =1$. This site points out the existence of Dirichlet's test when discussing towards this problem.
The test states that if
- $\left\{a_n\right\}$ is monotonically non-increasing
- $\lim_{n\rightarrow\infty} a_n=0$
- $\displaystyle\left|\sum_{1\le n\le N} b_n\right|\le M$ for all $N\ge 1;\exists M>0$
then the following converges, $\displaystyle\sum_{n\ge 1}a_nb_n$.
Is it okay to prove the test like this? $$\left|\sum_{1\le n\le N}a_nb_n\right|\le \sum_{1\le n\le N}|a_nb_n|\le |a_1|\sum_{1\le n\le N}|b_n|\le |a_1|M$$ Thus, on taking limit $n\rightarrow\infty$, the sum converges absolutely and hence converges $\blacksquare$
I, then, proceed with the $\sum_{n\ge 1} z^n/n$ by taking $a_n=1/n$ and $b_n=z^n$ in the test. We have $$\left|\sum_{1\le n\le N}z^n\right|\le M-1, $$ if $1+z+z^2+\cdots +z^{M-1}=0$ for some $M>0$, for $z\not =1$ so we are done.
Edited: So the proof is not true because of the previous to the last one.
No, that doesn't work. If it worked, it would follows that the series $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$ is absolutely convergent. But it isn't.
Note that you are assuming that, for some $M>0$, you have$$(\forall N\in\Bbb N):\left|\sum_{n=1}^Nb_n\right|\leqslant M.$$In your proof, you have used a much stronger assumption, which is$$(\forall N\in\Bbb N):\sum_{n=1}^N\left|b_n\right|\leqslant M.$$