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My intuition is that the first-order term in the Taylor expansion should dominate the series, if divergent:

$$ \log(1+x) = x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \frac { x ^ { 4 } } { 4 } + \cdots $$

So we would get with Taylor expansion:

$$ \sum_i\log(1+q_i) = \sum_{i}\sum_{k=1}^\infty\frac{q_i^k}{k} = \sum_i q_i + \sum_{i}\sum_{k=2}^\infty\frac{q_i^k}{k} $$

Is there a slick way to control the residual?

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  • $\begingroup$ If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples. $\endgroup$ Commented Jan 10, 2019 at 18:28
  • $\begingroup$ Can you please come up with one, very curious. $\endgroup$
    – nakajuice
    Commented Jan 10, 2019 at 18:42
  • $\begingroup$ I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = \log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $\sum_i(q_i-r_i)=\infty$ while $\sum_i r_i$ converge. $\endgroup$ Commented Jan 10, 2019 at 18:48
  • $\begingroup$ @MathLover nice, thank you $\endgroup$
    – nakajuice
    Commented Jan 10, 2019 at 18:56

3 Answers 3

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Hint:

Consider $q_0=1$, $q_1=-1/2$, and $q_{i+2}=q_i$ for $i=0,1,2,\cdots$

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For $q_i>0$ or equivalently $\ln (1+q_i)>0$, each series fails to equal $\infty$ iff it has a finite limit instead. This is where it helps to work with the contrapositive. If $\sum_i\ln (1+q_i)$ is finite for an infinite sequence $q_i$, $\lim_{i\to\infty}\ln (1+q_i)=0$ so $\lim_{i\to\infty}q_i=0$. Since $\ln (1+q_i)\sim q_i$ for small $q_i$, this implies $\sum_i q_i$ also converges. A comment of Sangchui Lee's references the fact that, with a suitable choice of not-all-positive $q_i$ (viz MathLover's answer), the sum of logarithms might approach neither $\infty$ nor $-\infty$. (Their example gives partial sums of the $\ln (1+q_i)$ equal to either $\ln 2$ or $0$, so there's no limit, infinite or otherwise.)

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User @Math Lover provided an example where $ \sum_{n=0}^{\infty} q_n = \infty$ but $ \sum_{n=0}^{N} \log(1+q_n)$ alternates between $\log 2$ and $0$ in $N$. In this answer, we construct an example where

$$ \sum_{n=0}^{\infty} q_n = \infty \quad \text{and} \quad \sum_{n=0}^{\infty} \log(1+q_n) = 0.$$

To this end, we prepare two auxiliary sequences:

  • $(\epsilon_k)_{k=1}^{\infty}$ is a sequence such that $1 \geq \epsilon_k \geq \epsilon_{k+1} > 0$ and $\epsilon_k \to 0$ as $k \to \infty$.
  • $(N_k)_{k=1}^{\infty}$ is a sequence of positive integers such that $ N_k \epsilon_k^2 \geq 1$ for all $k$.

Now define $(r_n)$ and $(q_n)$ by

$$ (r_n)_{n=0}^{\infty} = ( \underbrace{ \epsilon_1, -\epsilon_1, \cdots, \epsilon_1, -\epsilon_1 }_{2N_1\text{-terms}}, \underbrace{ \epsilon_2, -\epsilon_2, \cdots, \epsilon_2, -\epsilon_2 }_{2N_2\text{-terms}}, \cdots ), \qquad q_n = e^{r_n} - 1. $$

Then $r_n = \log(1+q_n)$, and

  1. By the alternating series test, $\sum_{n=0}^{\infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.

  2. Using the fact that $e^x \geq 1 + x + \frac{1}{e}x^2$ for $|x| \leq 1$, it follows that

    $$ q_n \geq r_n + \frac{1}{e}r_n^2 $$

    Now summing over $n$, we obtain

    $$ \sum_{n=0}^{\infty} q_n \geq \left( \sum_{n=0}^{\infty} r_n \right) + \left( \sum_{k=1}^{\infty} 2N_k \cdot \frac{\epsilon_k^2}{e} \right) \geq \sum_{k=1}^{\infty} \frac{2}{e} = \infty. $$

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