Let $a_n\geq0$ be a sequence such that for every sequence $b_n$ that converges to $0$, $\sum_{n=1}^\infty a_nb_n$ converges. Show that $\sum_{n=1}^\infty a_n$ converges.

My try:

I decided to show the negation of the statement: if $a_n\geq0$ is a sequence such that $\sum_{n=1}^\infty a_n$ diverges, then there is a sequence $b_n\rightarrow0$ such that $\sum_{n=1}^\infty a_nb_n$ diverges.

Since $a_n\geq0$, the divergence of the sum $\sum_{n=1}^\infty a_n$ implies that the sequence of partial sums, $S_n=\sum_{j=1}^n a_j$ is not bounded.

So for every M there is an N such that for every $n>N$ we get $S_n>M$.

This is as far as I got, since trying to construct a fitting $b_n$ worked in my head for certain examples, but I couldn't find a general construction. I can say really informaly that $b_n$ should be a sequence "weaker" than $a_n$, like in the following case:

$a_n=\frac1n$ then $b_n$ can be $\frac1{\ln{n}}$

but how about this one?

$a_n=\frac1{n\ln{n}}$ then $b_n=?$

I would appreciate directions of thought.


Let $a_n$ be such that $$\sum_{n=1}^{\infty}a_n=+\infty$$ Suppose, without loss of generality, $a_1>0$ (otherwise you can start from the first non null $a_n$). For $n\in\mathbb{N}^+$, let $$S(n)=\sum_{i=1}^na_n$$ By hypothesis, $S(n)$ diverges. Now $S(n)$ is strictly positive, so we can define $$b(n)=\frac{1}{\sqrt{S(n)}}$$ Since $a_n$ are positive and $S(n)$ diverges, $b(n)$ is positive, decreasing and converges to $0$. Moreover $$\sum_{n=1}^Na_nb_n\geq b_N\sum_{n=1}^Na_n=\sqrt{S(N)}$$ Hence $$\sum_{n=1}^{\infty}a_nb_n=+\infty$$

  • $\begingroup$ Awesome, thanks! $\endgroup$ – Whyka Oct 3 '15 at 1:28
  • $\begingroup$ You're welcome! $\endgroup$ – Capublanca Oct 3 '15 at 1:28

Another take:

If $\sum_n a_n$ diverges, the partial sums are not Cauchy. In particular, there is some $\epsilon>0$ such that for all $N$, there are $n >m\ge N$ such that $a_{m+1}+\cdots+a_n \ge \epsilon$.

By repeating (take $N' = n+1$ and repeat) we obtain a collection of finite (not overlapping) sequences such that $a_{m_k+1} + \cdots + a_{n_k} \ge \epsilon$ (with $n_k < m_{m_{k+1}+1}$ for all $k$).

Now define $b_{m_k+1} = \cdots = b_{n_k} = {1 \over k}$, for all $k$, with the remaining $b_i = 0$.

Then we see that $b_k \to 0$, but $\sum_{k=1}^{n_k} a_k b_k \ge \epsilon(1+ {1 \over 2} + \cdots + {1 \over k}) $, and so $\sum_n a_n b_n$ diverges.


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