# $\sum_{n=1}^\infty\ln(1+u_n)$ converges but $\sum_{n=1}^\infty u_n$ diverges

We know that if $$\sum\limits_{n=1}^\infty u_n^2<+\infty$$ is convergent, then both $$\sum\limits_{n=1}^\infty u_n$$ and $$\sum\limits_{n=1}^\infty \ln(1+u_n)$$ converge or diverge simultaneously. If $$\sum\limits_{n=1}^\infty u_n$$ converges and $$\sum\limits_{n=1}^\infty u^2_n=+\infty$$, then $$\sum\limits_{n=1}^\infty \ln(1+u_n)=-\infty$$；If $$\sum\limits_{n=1}^\infty \ln(1+u_n)$$ converges and $$\sum\limits_{n=1}^\infty u^2_n=+\infty$$, then $$\sum\limits_{n=1}^\infty u_n=+\infty$$.

My question is that: can we construct a sequence $$\{u_n\}$$ such that $$\sum\limits_{n=1}^\infty\ln(1+u_n)$$ is convergent but $$\sum\limits_{n=1}^\infty u_n$$ is divergent? In fact, it is suffices to construct $$\{u_n\}$$ such that $$\sum_{n=1}^\infty u_n=+\infty,\quad \sum_{n=1}^\infty \frac{u^2_n}{2}=+\infty,\quad \sum_{n=1}^\infty|u_n|^3<+\infty$$ and $$\sum\limits_{n=1}^\infty\left(u_n-\frac{u^2_n}{2}\right) \text{is convergent.}$$ Does such a sequence exist？

• @BrianMoehring here $u_n$ is alternated sequence. does it exists?
– HGF
Commented May 12, 2023 at 8:08

Take $$u_n = \exp \left( {\frac{{( - 1)^n }}{{\sqrt n }}} \right) - 1.$$ Then $$\sum\limits_{n = 1}^\infty {\log (1 + u_n )} = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{\sqrt n }}}$$ converges, but $$\sum\limits_{n = 1}^\infty {u_n } = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{\sqrt n }}} + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{1}{n}} + \mathcal{O}(1)\sum\limits_{n = 1}^\infty {\frac{{1 }}{{n^{3/2} }}}$$ diverges since the harmonic series diverges. (I used the fact that $$\exp(x)=1+x+\frac{x^2}{2}+\mathcal{O}(x^3)$$ for small $$x$$.)

• A great answer. By the way, $(e^x-1)^3=x^3+O(x^4)$, $e^x-1-\frac{1}{2}(e^x-1)^2=x-x^3/3+O(x^4)$ for small $x$. Hence $\sum_{n=1}^\infty u_n=+\infty,\quad \sum_{n=1}^\infty \frac{u^2_n}{2}=+\infty,\quad \sum_{n=1}^\infty|u_n|^3<+\infty$ and $\sum\limits_{n=1}^\infty\left(u_n-\frac{u^2_n}{2}\right)$ is convergent.
– HGF
Commented May 12, 2023 at 13:43