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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?

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7 Answers 7

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

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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.

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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 $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{b_n}=\infty$.

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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.

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

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  • $\begingroup$ (+1) I like the simplicity of this approach. $\endgroup$
    – robjohn
    Commented May 2, 2019 at 19:17
  • $\begingroup$ Nice+. An educational version: Assume that $\sum_{k} \frac{a_k}{b_k}<\infty$. Then $\frac{1}{2}>\sum_{k>=n} \frac{a_k}{b_k}$ for some $n$ - by the assumption. Fix this $n$. Then $\sum_{k>=n} \frac{a_k}{b_k}\ge \frac{1}{b_n}\sum_{k>=n} a_k=\frac{1}{b_n}b_n=1$, by term by term comparison. Contradiction by $\frac{1}{2} > 1$. $\endgroup$ Commented Aug 14, 2020 at 11:19
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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?

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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 :

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

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  • $\begingroup$ please explain $\displaystyle \frac{b_n-b_{n+1}}{b_n} \ge \int_{b_{n+1}}^{b_n} \frac{dx}{x}$ $\endgroup$
    – Mathronaut
    Commented Aug 12, 2014 at 16:58
  • $\begingroup$ For any function $f$ positive on $[a,b], a < b $, we have (just make a quick drawing) : $$(b-a) \cdot \underset{x\in [a,b]}{\min} f(x) \le \int_a^b f(t) dt \le (b-a)\cdot \underset{x\in [a,b]}{\max} f(x) $$ Another way to see it is that the mean value of the function $f$ over $[a,b]$ is between the minimum and the maximum value of $f$ on this interval. $\endgroup$
    – yultan
    Commented Aug 12, 2014 at 19:48
  • $\begingroup$ Yes, but $x \leq b_n$ and thus $\frac{1}{x} \geq \frac{1}{b_n}$, but you need $\frac{1}{x} \leq \frac{1}{b_n}$. $\endgroup$
    – PhoemueX
    Commented Aug 13, 2014 at 8:14
  • $\begingroup$ Sorry I made a mistake. You are right it should be with $b_{n+1}$. Initial answer corrected, though. $\endgroup$
    – yultan
    Commented Aug 13, 2014 at 11:56
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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 $$ \begin{align} \sum_{k=m}^n\frac{a_k}{b_k} &=\sum_{k=m}^nu_k\\ &\ge\frac1{2\log(2)}\sum_{k=m}^n\log\left(\frac1{1-u_k}\right)\\ &=\frac1{2\log(2)}\sum_{k=m}^n\log\left(\frac{b_k}{b_{k+1}}\right)\\ &=\frac1{2\log(2)}\log\left(\frac{b_m}{b_{n+1}}\right)\tag4 \end{align} $$ 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.

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