TITLE: How prove $\sum_{n=1}^{\infty}\dfrac{a_{n}}{\ln{(1+n)}}<+\infty$
QUESTION [8 upvotes]: Question:

Define sequence $\{a_{n}\}$ such  $$0<a_{n}<1,n\in N^{+}$$ such
  $$\sum_{n=1}^{\infty}\dfrac{a_{n}}{\ln{(a_{n})}}$$ is convergent.
Show that
  $$\sum_{n=1}^{\infty}\dfrac{a_{n}}{\ln{(1+n)}}<+\infty.$$

My idea:Use
$$\ln{(1+x)}>\dfrac{x}{1+x}$$
$$0<a_{n}<1$$ 
so
$$\dfrac{a_{n}}{\ln{(1+n)}}<\dfrac{(1+n)a_{n}}{n}$$
then I can't continue.Thank you.

REPLY [7 votes]: Consider two sets: $A=\left\{n\in\mathbb{N}: a_n<\dfrac{1}{(n+1)^2}\right\}$, $B=\left\{n\in\mathbb{N}: a_n\ge\dfrac{1}{(n+1)^2}\right\}$.
Note that if $n\in B$, then $$|\ln a_n|\le\ln (n+1)^2=2\ln(n+1),$$
$$\ln(n+1)\ge\dfrac{|\ln a_n|}{2}.$$ Then
$$
\sum\limits_{n=1}^\infty \dfrac{a_n}{\ln(1+n)} ~~ = ~~ 
\sum\limits_{n\in A} \dfrac{a_n}{\ln(1+n)} + \sum\limits_{n\in B} \dfrac{a_n}{\ln(1+n)} \\
\le \sum\limits_{n\in A} \dfrac{1}{(n+1)^2\ln(1+n)} + \sum\limits_{n\in B} \dfrac{2a_n}{|\ln a_n|} ~~ \le  ~~ S_1 + 2S_2,
$$
where
$$
S_1 = \sum\limits_{n=1}^\infty \dfrac{1}{(n+1)^2 \ln(n+1)} <\infty,
$$
$$
S_2 = -\sum\limits_{n=1}^\infty \dfrac{a_n}{\ln a_n} < \infty. \quad (S_2\mbox{ is positive}).
$$