Find a sequence $(x_n)$ such that $(x_n)$ is monotonic, $\lim x_n=0$, $\sum_{n=1}^{\infty} ( 1-\frac{x_n}{x_{n+1}})$ converges 
Find a sequence $(x_n)$ satisfying these conditions:

*

*$(x_n)$ is monotonic and $\lim x_n=0$


*$\displaystyle \sum_{n=1}^{\infty} \left( 1-\frac{x_n}{x_{n+1}}\right)$ converges.

This problem just came after I had solved another one. That is "if $(x_n)$ is monotonic and $\lim x_n=a>0$, then $\displaystyle \sum_{n=1}^{\infty} \left( 1-\frac{x_n}{x_{n+1}}\right)$ converges". This one can be solved by using Cauchy criterion. The case $a<0$ is similar. I just wonder how things will be when $a=0$. A trivial condition is $\lim \frac{x_n}{x_{n+1}}=1$, but I cannot continue.
Thanks so much.
 A: Let $a_n=1-\frac{x_n}{x_{n+1}}$. Then we can reconstruct $(x_n)$ by the recursion $x_{n+1}=\frac{x_n}{1-a_n}$. We clearly have $a_n\to 0$ from the convergense of $\sum a_n$. For $x_n\to 0$, we need hence $a_n<0$ for almost all $n$. Why not try $a_n=-\frac1{n^2}$? Then $x_{n+1}=\frac{x_n}{1+\frac1{n^2}}=\frac{n^2}{n^2+1}x_n$. Unfortunately, this does not give us $x_n\to 0$.

Proposition.  Let $(x_n)$ be a sequence of nonzero reals. Let $a_n=1-\frac{x_n}{x_{n+1}}$ and assume that the series $\sum a_n$ converges and that $a_n<0$ for almost all $n$. Then $x_n$ converges to a strictly positive limit.
Proof. Pick $N$ so that $-1<a_n<0$ for all $n>N$.
From $\frac{x_n}{x_{nü+1}}=1-a_n$, we find for $m>n>N$ that $$\frac{x_n}{x_m}=(1-a_n)(1-a_{n+1})\cdots (1-a_{m-1}),$$ hence
$$\ln x_n-\ln x_m = \sum_{n\le k<m}\ln(1-a_k).  $$
For $0<t<1$, we have $\frac t2\le \ln (1+t)\le t$, hence
$$\tag1-\frac12\sum_{n\le k<m}a_k\le \ln x_n-\ln x_m\le - \sum_{n\le k<m}a_k.  $$
Since $\sum a_k$, converges, the partial sums are a Cauchy sequence, so by $(1)$, $(\ln x_n)$ is also Cauchy, hence has a real limit $y$. Then $\lim x_n=e^y>0$. $_\square$
