Dedekind and Bois-Reymond criterion 1)Dedekind criterion
If $\lim a_n=0$, $\sum_{n=1}^{\infty} (a_n-a_{n-1})$ converges absolutely and the partial sums of $\sum_{n=1}^{\infty}  z_n$ are bounded, then $\sum_{n=1}^{\infty} a_nz_n$ converges.
2)Bois-Reymond criterion
If $\sum_{n=1}^{\infty} (a_n-a_{n-1})$ converges absolutely and $\sum_{n=1}^{\infty}  z_n$ converges, then $\sum_{n=1}^{\infty} a_nz_n$ converges.
I am trying to prove these two theorems but I don't know how to, I would appreciate any hints or suggestions.
For 1): Let $Z_N=z_1+...+z_N$, one can express $$\sum_{n=1}^N a_nz_n=a_NZ_N-\sum_{n=0}^{N-1} Z_n(a_{n+1}-a_n)=a_NZ_N+\sum_{n=0}^{N-1} Z_n(a_n-a_{n+1})$$
From the inequality $0\leq |a_NZ_N|\leq |a_N|M$, one can deduce that $a_NZ_N \to 0$ when $N \to \infty$,
With respect to $\sum_{n=0}^{N-1} Z_n(a_n-a_{n+1})$:  $$|  \sum_{n=0}^{N-1} Z_n(a_n-a_{n+1})|\leq \sum_{n=0}^{N-1} |Z_n||(a_n-a_{n+1})|\leq M\sum_{n=0}^{N-1} |(a_n-a_{n+1})|$$, letting $N \to \infty$, the right member of the inequality converges.
 A: Your proof of the Dedekind criterion is correct. The proof for the du Bois-Reymond criterion must look different, since the proof of the Dedekind criterion depends on the assumption that $a_n \to 0$, which is not part of the hypotheses of the du Bois-Reymond criterion. Instead, we have the assumption that $\sum z_n$ converges, not only that it has bounded partial sums.
In such a case, instead of writing
$$z_n = \left(\sum_{k=1}^n z_k\right) - \left(\sum_{k=1}^{n-1} z_k\right),$$
one can also write
$$z_n = \left(\sum_{k=n}^\infty z_k\right) - \left(\sum_{k=n+1}^\infty z_k\right),$$
where now
$$R_n := \sum_{k=n}^\infty z_k \xrightarrow{n\to\infty} 0$$
can be used to make the solitary term(s) in a summation by parts small.
We find
$$\begin{align}
\sum_{k=1}^N a_k z_k &= \sum_{k=1}^N a_k(R_k - R_{k+1})\\
&= \sum_{k=1}^N a_k R_k - \sum_{k=2}^{N+1} a_{k-1}R_k\\
&= a_1R_1 + \sum_{k=2}^N (a_k - a_{k-1})R_k - a_NR_{N+1},
\end{align}$$
where the first term is constant, the sum is a partial sum of an absolutely convergent series (since the $R_k$ are bounded), and the final term converges to $0$, since $a_N$ is bounded and $R_{N+1} \to 0$. Together, the (conditional) convergence of
$$\sum_{k=1}^\infty a_k z_k$$
follows.
