# Does $\sum q_i = \infty$ imply $\sum \log(1+q_i)=\infty$?

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?

• If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples. Commented Jan 10, 2019 at 18:28
• Can you please come up with one, very curious. Commented Jan 10, 2019 at 18:42
• 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. Commented Jan 10, 2019 at 18:48
• @MathLover nice, thank you Commented Jan 10, 2019 at 18:56

Hint:

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

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.)

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

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