Let $\displaystyle \sum a_n$ be convergent series of positive terms and set $\displaystyle b_n=\sum_{k=n}^{\infty}a_k$ , then prove that $\displaystyle\sum \frac{a_n}{b_n}$ diverges.
I could see that $\{b_n\}$ is monotonically decreasing sequence converging to $0$ and I can write $\displaystyle\sum \frac{a_n}{b_n}=\sum\frac{b_n-b_{n+1}}{b_n}$, how shall I proceed further?
Best Answer
Sorry, my previous answer was not correct. A new tentative:
$$\frac{b_{k+1}}{b_{k}}=1-\frac{a_k}{b_k}$$ Hence $$\frac{b_{N+1}}{b_1}=\prod_{k=1}^N{(1-\frac{a_k}{b_k}})$$ and $$\log b_{N+1}-\log b_1=\sum_{k=1}^N \log(1-\frac{a_k}{b_k})$$ Now if the series $a_k/b_k$ is convergent, we have $a_k/b_k \to 0$, and as $\log(1-x)\sim -x$ and the series have constant sign, this imply that the series $\displaystyle \log (1-\frac{a_k}{b_k})$ is convergent, a contradiction as $\log (b_{N+1}) \to -\infty$.