Dedekind and Bois-Reymond criterion

Question

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.

I believe that the technique you are looking for is summation by parts. Take a look at http://en.wikipedia.org/wiki/Summation_by_parts. — jwsiegel, May 05 '14 at 02:38
Your work for Dedekind's criterion looks fine. Have you found an angle of attack for du Bois-Reymond? — Daniel Fischer, Jun 27 '14 at 20:36
Do you want to think for yourself some more, or a big hint now? — Daniel Fischer, Jun 27 '14 at 20:42

score 3 · Accepted Answer · answered Jun 28 '14 at 14:57

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.

Dedekind and Bois-Reymond criterion

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