Somemathhere

 Let $\{a_n\}_{n\ge 1}$ be a sequence of positive numbers such that  $$\sum_{n=1}^{\infty} a_n$$ is convergent. Then there exists a sequence $\{b_n\}_{n\ge 1}$ with $\displaystyle \lim_{n\to \infty}\frac{b_n}{a_n}=\infty$ and such that  $$\sum_{n=1}^{\infty} b_n$$ is also convergent. First proof (a la Calculus I, based on ideas of Zhivko Petrov ). Consider $$S_n=\sum_{k=n}^{\infty}a_k.$$ Clearly $S_n\to 0$. Define $$b_n=\sqrt{S_n}-\sqrt{S_{n+1}}.$$ Clearly $b_n>0$ for all $n$ and $b_n\to 0$. Moreover  $$\sum_{n=1}^\infty b_n=\sqrt{S_1},$$ hence is convergent. On the other hand $$\lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}\frac{1}{\sqrt{S_n}+\sqrt{S_{n+1}}}=\infty.$$   Second proof . (Based on Folland 's book) Assume on the contrary that there is lowest decaying sequence, i.e.  for some sequence $a=\{a_n\}_{n\ge 1}$ with convergent series, holds that for any sequence of positive numbers $\{b_n\}_{n\ge 1}$, the series $$\sum_{n=1}^\infty a_nb_n$$ i...