If $\sum a_n$ converges and $b_n=\sum\limits_{k=n}^{\infty}a_k $, prove that $\sum \frac{a_n}{b_n}$ diverges 
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?
 A: 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.
A: 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 $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{b_n}=\infty$.
A: 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?
A: 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. 
$$
\sum_{k=n}^{\infty}\frac{a_k}{b_k} \geq \frac{1}{b_n}\sum_{k=n}^{\infty}a_k = 1 = \epsilon
$$ 
A: 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 :
$$ \sum_{n=1}^N \frac{b_n-b_{n+1}}{b_{n+1}} \ge \sum_{n=1}^N \int_{b_{n+1}}^{b_n} \frac{dx}{x} = \int_{b_{N+1}}^{b_1} \frac{dx}{x} = \ln \left(\frac{b_1}{b_{N+1}}\right) \underset{N\to+\infty}{\longrightarrow}+\infty $$
A: 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 $\log (b_{N+1}) \to -\infty$. 
A: Preliminary
For $x\ge0$,
$$
\log\left(\frac1{1-x}\right)=\int_0^x\frac{\mathrm{d}t}{1-t}\le\frac{x}{1-x}\tag1
$$
therefore,
$$
\frac{\mathrm{d}}{\mathrm{d}x}\frac{\log\left(\frac1{1-x}\right)}x=\frac{\frac{x}{1-x}-\log\left(\frac1{1-x}\right)}{x^2}\ge0\tag2
$$
thus, $\frac{\log\left(\frac1{1-x}\right)}{x}$ is increasing on $(0,1)$. Thus, for $0\lt x\le\frac12$,
$$
\frac{\log\left(\frac1{1-x}\right)}{x}\le2\log(2)\tag3
$$

Application
Let $u_k=\frac{a_k}{b_k}$ and note that $\frac1{1-u_k}=\frac{b_k}{b_{k+1}}$. Assume the sum converges. Then for $k\ge m$, $u_k\le\frac12$, and then
$$
\begin{align}
\sum_{k=m}^n\frac{a_k}{b_k}
&=\sum_{k=m}^nu_k\\
&\ge\frac1{2\log(2)}\sum_{k=m}^n\log\left(\frac1{1-u_k}\right)\\
&=\frac1{2\log(2)}\sum_{k=m}^n\log\left(\frac{b_k}{b_{k+1}}\right)\\
&=\frac1{2\log(2)}\frac{b_m}{b_{n+1}}\tag4
\end{align}
$$
but the right side diverges since $\lim\limits_{n\to\infty}b_{n+1}=0$. Therefore, $\sum\limits_{k=1}^\infty\frac{a_k}{b_k}$ diverges.
