# Interesting fact about convergent sequences?

Let $a_n > 0$ for $n=1,2,...,$ with $\sum_{n=1}^{\infty}a_n < \infty$. Prove that $b_n$ $(n=1,2,...)$ exist such that $b_n/a_n \rightarrow \infty$ as $n \rightarrow \infty$, but $\sum_{n=1}^{\infty}b_n < \infty$.

This somehow is reminiscent of a criterion about sequences which says that, if $a_n$ and $b_n$ are non-negative with $b_n > a_n$, for $n=1,2,...,$ and $\sum b_n$ is convergent, then $\sum a_n$ must be convergent as well. How does one prove the statement above? I simply have no idea.

Any help appreciated.

Hint:

Let $A_n$ be the partial sum of $(a_n)$ and $A_n\to A$

Define $b_1=\sqrt{A}-\sqrt{A-A_1}$ and $b_n=\sqrt{A-A_{n-1}}-\sqrt{A-A_n}$

Then $$\frac{b_n}{a_n}=\frac{1}{\sqrt{A-A_{n-1}}+\sqrt{A-A_n}}\to \infty$$ but $\sum b_n<\infty$

Another solution:

For $k \in \mathbb{N}$, define $n_k:=\min\{ n\in \mathbb{N}: \sum_{j=n}^{\infty}a_j\leq 2^{-k}\}$. This gives a sequence of indices $n_1\leq n_2 \leq n_3 \leq\dots$ such that $2^{-(k+1)}<\sum_{j=n}^{\infty} a_j\leq 2^{-k}$ whenever $n_k\leq n<n_{k+1}$.

Define $b_n:= ka_n$ whenever $n_k \leq n <n_{k+1}$ (and define $b_n=0\;$ if $\;n<n_1$).

Then clearly $\frac{b_n}{a_n} \to \infty$, since $\frac{b_n}{a_n} \geq k$ whenever $n \geq n_k$.

However, we have that $$\sum_{n=n_1}^{\infty} b_n = \sum_{k=1}^{\infty} k\bigg[\sum_{n=n_k}^{n_{k+1}-1} a_n \bigg]\leq \sum_{k=1}^{\infty} k\bigg[\sum_{n=n_k}^{\infty} a_n \bigg]\leq \sum_{k=1}^{\infty} k2^{-k}<\infty$$