The problem is the following:

If $\sum \limits_{n=1} ^{\infty} a_n$ converges, where $a_n$ are real numbers, then there exists $b_n \to \infty$ so that $\sum \limits_{n=1} ^{\infty} a_n b_n$ is still convergent.

I know that the above statement is true if $a_n$ are nonnegative (setting $b_n=\frac1{\sqrt{R_{n-1}}+ \sqrt{R_{n}}}$, where $R_n=\sum \limits_{k=n} ^{\infty} a_k$). But for the general $a_n$, I have no idea how to prove it.

Any ideas are welcome. Thanks!


Take an increasing sequence $\{N_k\}$ such that $\left|\sum_{n=N}^M \ a_n\right| < 4^{-k}$ whenever $N_k \le N < M$, and let $b_n = 2^k$ for $N_k \le n < N_{k+1}$.

  • $\begingroup$ Nice solution. Thanks! $\endgroup$ – Syang Chen Aug 17 '11 at 14:31

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