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？
Best Answer
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 $x$.)