For Dirichlet's test, is it true that if $\ a_1=1\ $ then $\ \left\vert \sum_{n=1}^{\infty} a_n b_n \right\vert \leq M\ ?$ I think I might be gaining more understanding of Dirichlet's test visually.
For Dirichlet's test, is it true that if $\ a_1=1\ $ then $\ \displaystyle\left\vert \sum_{n=1}^{\infty}  a_n b_n \right\vert \leq M\ ?$
I think this because $\ \displaystyle\left\vert \sum_{n=1}^{k}\ b_n \right\vert\ \leq M\implies\ \sum_{n=1}^{k}\ b_n\in\ B(0+0i,M)=\{\ \vert z\vert\leq M:z\in\mathbb{C}\ \}\ $ for every $\ k.\ $
Therefore it seems to me that if $\ a_1=1\ $ and $\ \{a_n\}\ $ is decreasing with $\ a_n\to 0,\ $ then the terms of $\ \displaystyle\sum_{n=1}^{k}  a_n b_n\ $ ought to also be "bouncing around" inside $\ B(0+0i,M).\ $ Is this picture correct? And if so is there a proof of this? Or is the proof of this covered in the proof of Dirichlet's test?
 A: Let $T_n = \sum_{k=1}^{n} b_k$ be the partial sum of $(b_n)$. Then by summation by parts,
\begin{align*}
\sum_{k=1}^{n} a_k b_k
&= \sum_{k=1}^{n} a_k (T_k - T_{k-1}) \\
&= a_n T_n + \sum_{k=1}^{n-1} (a_k - a_{k+1}) T_k.
\end{align*}
If $T_k$ is bounded and $(a_k)$ decreases to $0$, then as $n \to \infty$ the above sum converges to
$$ \sum_{k=1}^{\infty} a_k b_k = \sum_{k=1}^{\infty} (a_k - a_{k+1}) T_k. $$
In particular, if $(T_n)$ is bounded by $M$, then
$$ \left| \sum_{k=1}^{\infty} a_k b_k \right| \leq M \sum_{k=1}^{\infty} (a_k - a_{k+1}) = M a_1. $$
This resolves OP's question.
A: Suppose $\ a_1=1,\ \{a_n\}\ $ is monotonically decreasing with $\ a_n\to 0,\ $ and suppose that $\ \displaystyle\left\vert\sum_{n=1}^k b_n\right\vert \leq M\quad \forall k\in\mathbb{N}.\ $ We can use the result of Dirichlet's Test also, that is, $\ \displaystyle\sum_{n=1}^{\infty} a_n b_n\ $ converges. I first aim to show that $\ \displaystyle\left\vert\sum_{n=1}^k a_n b_n\right\vert \leq M\quad \forall k\in\mathbb{N}.$
We can make use of the following identity, which holds for all $\ k\geq 2$:
$$\sum_{n=1}^k a_n b_n = \sum_{n=1}^{k-1} \left( (a_n-a_{n+1}) \sum_{j=1}^n b_j \right) + a_k\sum_{n=1}^k b_n.$$
For $\ k=1,\ $ we have $\ \left\vert\displaystyle\sum_{n=1}^k a_n b_n\right\vert = \vert a_1 b_1 \vert = a_1 \vert b_1 \vert \leq \vert b_1\vert \leq M.$
For every $\ k\geq2, $ we have:
\begin{align} \left\vert\sum_{n=1}^{k} a_n b_n\right\vert = \left\vert\sum_{n=1}^{k-1} \left( (a_n-a_{n+1}) \sum_{j=1}^n b_j \right) + a_k\sum_{n=1}^k b_n\right\vert\\
\\
\leq \sum_{n=1}^{k-1} \left( (a_n-a_{n+1}) \left\vert\sum_{j=1}^n b_j\right\vert \right) + a_k\left\vert\sum_{n=1}^k b_n \right\vert \\
\\
\leq \left( \sum_{n=1}^{k-1} (a_n-a_{n+1}) M\right) + a_n M = a_1 M \leq M.\\
\\
\end{align}
$$$$
This means that, for each $\ k\geq 1,\ \displaystyle\sum_{n=1}^{k} a_n b_n $ is a point inside the closed disc $\ \{\ \vert z\vert\leq M:z\in\mathbb{C}\ \}.\ $ Since $\ \displaystyle\sum_{n=1}^{\infty} a_n b_n\ $ converges, $\ \displaystyle\sum_{n=1}^{\infty} a_n b_n\ $ must converge to a member of $\ \{\ \vert z\vert\leq M:z\in\mathbb{C}\ \},\ $ which implies that $\ \displaystyle\left\vert \sum_{n=1}^{\infty}  a_n b_n \right\vert \leq M.$
