If $\sum a_n$ converges and $b_n=\sum\limits_{k=n}^{\infty}a_k $, prove that $\sum \frac{a_n}{b_n}$ diverges

Question

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?

Answer 1

For $m>n$ one has $$\begin{aligned}\frac{a_n}{b_n}+\cdots\frac{a_m}{b_m}&\ge \frac{a_n}{b_n}+\cdots +\frac{a_m}{b_n}\\ &=\frac{b_n-b_{m+1}}{b_n} = 1-\frac{b_{m+1}}{b_n}. \end{aligned}$$ Can you continue from here?

Answer 2

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 $\log (b_{N+1}) \to -\infty$.

Answer 3

First note that $\{b_n\}_{n\in\mathbb N}$ is a decreasing sequence of positive numbers, which tends to zero.

We have that, for $n> m$ $$ \frac{a_m}{b_m}+\cdots+\frac{a_n}{b_n}\ge \frac{a_m}{b_m}+\cdots+\frac{a_n}{b_m}=\frac{1}{b_m}(a_m+\cdots+a_n)=\frac{b_n-b_m}{b_m}=1-\frac{b_n}{b_m}. $$ Next, as $b_n\searrow 0$, choose $m_1,m_2,\ldots,m_k,\ldots$, so that $$ \frac{b_{m_{i+1}}}{b_{m_i}}<1/2. $$ Then we have that $$ \sum_{n=1}^{m_k}\frac{a_n}{b_n}\ge\sum_{i=1}^{k-1}\sum_{n=m_i+1}^{m_{i+1}}\frac{a_n}{b_n}\ge \sum_{i=1}^{k-1}\left(1-\frac{b_{m_i}}{b_{m_{i+1}}}\right)\ge\frac{k-1}{2}, $$ and hence $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{b_n}=\infty$.

Answer 4

Assuming $\sum a_n=L$, for any $n$ big enough we must have: $$a_n \geq \frac{1}{n}\sum_{m>n}a_m,\tag{1}$$ otherwise, assuming that $N$ is the greatest integer for which $(1)$ holds, we have: $$L< a_N+\frac{1}{N}(L-a_n)+\frac{N-1}{(N+1)N}(L-a_n)+\ldots = L,\tag{2}$$ contradiction. This implies that for any $n\geq M$ $$\left(1+\frac{1}{n}\right)a_n\geq\frac{1}{n}b_n\tag{3}$$ holds, hence: $$\sum_{n\geq M}\frac{a_n}{b_n}\geq\sum_{n\geq M}\frac{1}{n+1},\tag{4}$$ but the RHS of $(4)$ diverges.

Answer 5

For a sum $\sum_{k=0}^{\infty}c_k$ to converge its tail must converge to 0.

$$ \lim_{n \to \infty} \sum_{k=n}^{\infty}c_k = 0 $$

i.e. for every $\epsilon$ there exist a $n_0$ such that for all $n > n_0$

$$ \left| \sum_{k=n}^{\infty}c_k \right| < \epsilon $$

We can prove that there exists a $\epsilon$ such that for all $n$ the sum is bigger than $\epsilon$ and thus the sum diverges.

$$ \sum_{k=n}^{\infty}\frac{a_k}{b_k} \geq \frac{1}{b_n}\sum_{k=n}^{\infty}a_k = 1 = \epsilon $$

Answer 6

First answer was wrong. Here is my new try :

If $\underset{n\to+\infty}{\lim} \frac{a_n}{b_n}$ is not $ 0 $ or does not exist, $\sum \frac{a_n}{b_n}$ is directly divergent.

Otherwise we have $\underset{n\to+\infty}{\lim} \frac{a_n}{b_n} = 0 $ so $b_{n+1} \sim b_n $ . It implies $\sum \frac{a_n}{b_n} = \sum \frac{b_n-b_{n+1}}{b_n}$ and $\sum \frac{b_n-b_{n+1}}{b_{n+1}}$ have the same behavior thanks to limit comparison test.

Now, since $b_n$ is decreasing and tends to $0$ as $n\to+\infty$, we can use an integral comparison :

$$ \sum_{n=1}^N \frac{b_n-b_{n+1}}{b_{n+1}} \ge \sum_{n=1}^N \int_{b_{n+1}}^{b_n} \frac{dx}{x} = \int_{b_{N+1}}^{b_1} \frac{dx}{x} = \ln \left(\frac{b_1}{b_{N+1}}\right) \underset{N\to+\infty}{\longrightarrow}+\infty $$