Prove that there exists a decreasing sequence $b_n \rightarrow 0$ such that $\sum_{n=1}^{\infty} a_n b_n$ diverges.

Question

It is given that $a_n>0$ and $\sum_{n=1}^{\infty} a_n$ is divergent. Prove that there exists a decreasing sequence $b_n \rightarrow 0$ such that $\sum_{n=1}^{\infty} a_n b_n$ diverges.

I thought of an example: Let $b_1=1$ and $b_{n+1}=\min (1/(2^n a_{n+1}), b_n, 1/n)$ for $n \geq 1$. Will this work?

Answer 1

Since $\sum_{n \geq 1}a_n = +\infty$, there is $N_1 \geq 1$ such that we have $$a_1+\cdots+ a_{N_1} \geq 1.$$Then define $b_1 = \cdots = b_{N_1}= 1$. Then, since $\sum_{n > N_1}a_n = +\infty$, we get $N_2 > N_1$ such that $$a_{N_1+1}+\cdots+ a_{N_2} \geq 2,$$and we put $b_{N_1+1}=\cdots = b_{N_2} = 1/2$. Similarly, we get $N_3 > N_2$ such that $$a_{N_2+1}+\cdots + a_{N_3} \geq 3,$$and we put $b_{N_2+1}=\cdots = b_{N_3} = 1/3$. Keep going on. By construction, $b_n \downarrow 0$, and we have $$\begin{align}\sum_{n\geq 1}a_nb_n &= \sum_{k \geq 1}\sum_{n=N_k}^{N_{k+1}}a_nb_n = \sum_{k \geq 1}\sum_{n=N_k}^{N_{k+1}}\frac{a_n}{k} \geq \sum_{k \geq 1}\sum_{n=N_k}^{N_{k+1}} \frac{k}{k} \\ &= \sum_{k \geq 1} (N_{k+1}-N_k) = \lim_{k \to +\infty} N_{k+1}-N_1 = +\infty. \end{align}$$

Answer 2

Let consider for example

• $a_n =\frac1n$

• $b_n =\frac1{\log n}$

and $\sum \frac1{n\log n}$ diverges by Cauchy condensation test.