Let $\{a_n\}$ be a decreasing sequence of non-negative real numbers. Suppose that $$\lim_{n\to \infty} a_n = 0$$ Also, assume that the partial sum sequence $\{B_n\}$ of the series $\sum_{n=1}^\infty b_n $ is bounded. Show that the series $\sum_{n=1}^\infty a_nb_n$ converges.

My attempt:

To show that the series $\sum_{n=1}^\infty a_nb_n$ converges, I tried to show that sequence of partial sums $\{\sum_{i=1}^n a_ib_i\}_{n\in \mathbb{N}}$ converges. I tried to do this by showing that this sequence of partial sums is Cauchy.

Let $\varepsilon > 0$. Let $M$ be the bound on $\{B_n\}$. Since $\displaystyle\lim_{n\to \infty} a_n = 0$ we have that there exists $N\in \mathbb{N}$ such that for all $n > N$ we have $|a_n| < \dfrac{\varepsilon}{2M}$. Let $m,n > N$. We have that \begin{align*} \left|\sum_{i=1}^n a_ib_i - \sum_{i=1}^m a_ib_i\right| &= \left|\sum_{i=1}^N a_i b_i - \sum_{i=N+1}^n a_i b_i -\sum_{i=1}^N a_i b_i + \sum_{i=N+1}^m a_ib_i \right|\\ &=\left| \sum_{i=N+1}^m a_ib_i - \sum_{i=N+1}^n a_i b_i\right|\\ &\leq \left| \sum_{i=N+1}^m a_ib_i\right| + \left|\sum_{i=N+1}^n a_i b_i\right|\\ &\leq \text{ something }\\ &\leq |a_N|M + |a_N|M \\ &<\varepsilon \end{align*}

My problem is the middle step/s labeled "something". I'm not sure how to relate the bound $M$ to $\sum_{i=N+1}^n a_i b_i$. I'd appreciate any help. Is this the right idea for the problem?

Thanks in advance!

  • $\begingroup$ Have you heard of Abel's transformation? $\endgroup$ – codetalker May 10 '17 at 18:21

There is a formula called 'partial summation' formula or Abel's transformation which is useful in this case. It basically states that:

$$\sum_{i=1}^na_nb_n=\sum_{i=1}^{n-1}A_n(b_n-b_{n+1})+A_nb_n$$ where $$A_n=\sum_{i=1}^na_n$$

You have $$\lim_{n\to\infty}b_n=0$$such that $$b_n>b_{n+1},\forall n$$ and that $$|A_n|\le M$$for some $M\in\mathbb{R}$

If you apply the formula to the series, then notice that $$A_nb_n\to0$$ And, you have $$|A_n(b_n-b_{n+1})|\le M(b_n-b_{n+1})$$ so that the series $$\sum_1^\infty A_n(b_n-b_{n+1})$$ is convergent(the series of terms on the right side of the inequality is telescoping,so convergent. Hence the comparison test is used). Your claim now follows.

  • $\begingroup$ I have $\lim_{n\to \infty} a_n = 0$, not $b_n$. But yeah that seems to work out if I switch the notation around. $\endgroup$ – M47145 May 10 '17 at 20:20

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