For Dirichlet's test, is it true that if $\ a_1=1\ $ then $\ \left\vert \sum_{n=1}^{\infty} a_n b_n \right\vert \leq M\ ?$

Solution 1:

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.

Solution 2:

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 $\ \displaystyle\left\vert \sum_{n=1}^{\infty} a_n b_n \right\vert \leq M.$