Dirichlet's/Abel's test like convergence to $0$ for absolutely convergent series

Question

Consider the series $S_N = \sum_{k=0}^N a_k b_k$. Let:

• $a_k > 0$
• $\sum_{k=0}^N a_k < M$ for any $N$
• $a_k \rightarrow 0$ as $N \rightarrow \infty$
• $b_k = \beta^{N-k}$ with $\beta \in (0,1)$.

I am pretty sure that this implies convergence of $\sum_{k=0}^N a_k b_k$ to $0$ as $N \rightarrow \infty$, because $a_k$ is finite and $b_k$ tends to $0$ for increasing $N$. But unfortunately, I'm not able to prove it.

The problem is that all available tests/theorems aim to prove convergence to a finite value instead of to zero. I tried a Abel's test/Dirichlet's test like approach (with summation of parts) but this alsways ends up in convergence to a nonzero value.

Any help and hint is appreciated. Thanks!

Answer 1

Just write $$\sum_{k=0}^N a_kb_k = \sum_{k=0}^N a_k \beta^{N-k}=\frac{\sum_{k=0}^N a_k \beta^{-k}}{\beta^{-N}}$$

and apply Kronecker's lemma.

Answer 2

Welcome in MSE.

This is tricky! The main tool we use is the summation by parts, which is an analogue of the integration by parts in the discrete world. Setting

$$T_n := \sum_{k=0}^n \frac{a_k}{\beta^k} $$

You are asking whether $\beta^n T_n$ tends to zero. Notice that $(T_p - T_{p-1})\beta^p = a_p$. Summing from $p=0$ to $p=N$ and using the second condition we have

$$ M > \sum_{p=0}^N (T_p - T_{p-1}) \beta^p $$

And applying summation by parts

$$ M > \beta^N T_{N+1} - T_0 - \sum_{p=1}^N T_p (\beta^p - \beta^{p-1} ) = \beta^n T_{n+1} - T_0 + (1-\beta)\sum_{p=1}^N T_p \beta^{p-1} $$

$$ > - T_0 + (1-\beta) \sum_{p=1}^{N+1} T_p \beta^{p-1} $$

Which implies

$$ \frac{\beta(M+T_0)}{1-\beta} > \sum_{p=1}^{N+1} T_p \beta^p $$

Now the right hand side is a bounded and positive series in $N$, hence it will converge for $N \to \infty$. Since it converges, the general term $T_p \beta^p$ must converge to zero, which was our thesis!