Does the series $\sum\frac{a_n}{a_{n+1}}\frac{1}{n}$ always diverge if $0Does the series
$$\sum_{n=1}^{\infty}\frac{a_n}{a_{n+1}}\frac{1}{n}$$
diverge for any $a_n$, satisfying $0<a_n<1$, $n=1, 2, 3\dots$ ?
 A: For each $n$, let $b_n\in (-\infty,0)$ such that $\exp(b_n)=a_n$. By convexity of the exponential function on $\mathbb R$, 
$$\tag{1}\exp\left(\frac 1M\sum_{j=1}^M(b_j-b_{j+1})\right)\leqslant \frac 1M\sum_{j=1}^M\exp(b_j-b_{j+1})=\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}.$$
Assume that $\sum_{j=1}^\infty\frac{a_j}{a_{j+1}}\frac 1j$ is convergent. Since for each $N\leqslant M$, 
$$\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}\leqslant 
\frac 1M\sum_{j=1}^N\frac{a_j}{a_{j+1}}+\sum_{j=N}^M\frac{a_j}{a_{j+1}}\underbrace{\frac 1M}_{\leqslant \frac 1j}\leqslant\frac 1M\sum_{j=1}^N\frac{a_j}{a_{j+1}}+\sum_{j=N}^M\frac{a_j}{a_{j+1}}\frac 1j,$$
we obtain that 
$$\limsup_{M\to +\infty}\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}\leqslant\sum_{j=N}^\infty\frac{a_j}{a_{j+1}}\frac 1j,$$
hence by (1), 
$$\lim_{M\to +\infty}\exp\left(\frac 1M\sum_{j=1}^M(b_j-b_{j+1})\right)=0.$$
Since $\sum_{j=1}^M(b_j-b_{j+1})=b_1-b_{M+1}$, it follows that 
$$\lim_{M\to +\infty}\exp(-b_{M+1}/M)=0$$
which is not possible because each $b_M$ is negative.
The argument would boil down if we had considered $\sum_{n=1}^\infty\frac{a_{n+1}}{na_n}$ and in this case the series may converge, for example with $a_n:=2^{-n^2}$.
