I want to show the following result:
If $(a_n)$ and $(b_n)$ are monotone sequences with limit $0$, then the Cauchy's product series of $\sum(-1)^na_n$ and $\sum(-1)^nb_n$ is convergent if, and only if, the numbers $\sigma_n = a_n(b_0+b_1+\ldots+b_n)$ and $\tau_n = b_n(a_0+a_1+\ldots+a_n)$ also form a null sequence.
I think that this result is quite interesting: for example we can infer that the Cauchy's product series of two alternating harmonic series is convergent.
We can assume both $a_n, b_n \ge 0$ monotone descending. It's easy to show that those conditions are necessary, by just looking at the general term $c_n$ of the product series: $$c_n = (-1)^n(a_0b_n+a_1b_{n-1}+\ldots + a_nb_0)$$ for which the absolute value can be lower bounded by both $\sigma_n$ and $\tau_n$ (and hence cannot tend to $0$ if $\limsup \sigma_n > 0$ or $\limsup \tau_n > 0$). I don't really know how to prove that these conditions are also sufficient. Can you just give me a hint? Thanks in advance!
Best Answer
I'll assume that $a_n$ and $b_n$ are nonnegative, otherwise we just have to multiply everything with $-1$ once or twice. Further I'll write $A_n = a_0 + \ldots + a_n$ and similarly for $B_n$, and deviating from your notation I'll write $$c_n = a_0b_n + \ldots + a_nb_0\,.$$ Thus the Cauchy product of the two series $\sum (-1)^na_n$ and $\sum (-1)^nb_n$ is $\sum (-1)^nc_n$.
Now trivially a necessary condition for the convergence of the Cauchy product is $c_n \to 0$. Since $$\sigma_n = a_nB_n \leqslant c_n \qquad\text{and}\qquad \tau_n = A_nb_n \leqslant c_n$$ it is immediate that convergence of the Cauchy product implies $\sigma_n \to 0$ and $\tau_n \to 0$. In the other direction, we first observe that splitting the sum defining $c_n$ in the middle yields the inequality $$c_n \leqslant a_{\lfloor n/2\rfloor} B_{\lfloor n/2\rfloor} + A_{\lfloor n/2\rfloor} b_{\lfloor n/2\rfloor}\,.$$ Hence $\sigma_n \to 0$ and $\tau_n \to 0$ together imply $c_n \to 0$. If we could show that $c_n$ is eventually decreasing, convergence of the Cauchy product would then follow by the Leibniz criterion. But I don't see how that could be shown, so I'll take a less direct route.
It is clear that the sequence of products of the partial sums, $$P_n = \Biggl(\sum_{k = 0}^{n} (-1)^k a_k\Biggr)\Biggl(\sum_{k = 0}^{n} (-1)^k b_k\Biggr)\,,$$ converges, and therefore it is sufficient to show that $$P_n - \sum_{k = 0}^{n} (-1)^k c_k$$ tends to $0$. We can write this difference as $$(-1)^{n+1} \sum_{j = 1}^{n} (-1)^{j-1} s_{n,j}\,,$$ where $$s_{n,j} = \sum_{m = j}^{n} a_mb_{n+j-m}\,.$$ For $1 \leqslant j < n$ we have $$s_{n,j+1} = \sum_{m = j+1}^{n} a_mb_{n+j+1-m} \leqslant \sum_{m = j+1}^{n} a_m b_{n+j-m} \leqslant s_{n,j}$$ by monotonicity of $(b_n)$ and nonnegativity, whence $$\Biggl\lvert\, P_n - \sum_{k = 0}^{n} (-1)^k c_k\Biggr\rvert \leqslant s_{n,1} \leqslant c_{n+1} \to 0\,.$$