For any divergent series $ \sum x_n$, there exists a series $\sum a_n$ such that $\limsup \frac{a_n}{x_n}=0,$ but still $\sum a_n$ diverges. Q. Show that for any divergent series $ \sum x_n$, there exists a series $\sum a_n$ such that $$\limsup \frac{a_n}{x_n}=0,$$ but still $\sum a_n$ diverges.
My attempt:
For the sake of contradiction, Suppose that $\sum \frac{x_n}{1+x_n}$ converges, then $x_n >0$, implies $\frac{x_n}{1+x_n}=\frac{1}{1+1/x_n} \to 0$ and thereby $x_n \to 0$. Now, due to the fact $\frac{x_n}{1+x_n}>\frac{1}{2}x_n$ whenever $x_n<1$, we are in a contradiction. But I am confused that how to apply the condition $x_n>0$ for limit supremum??
 A: Let $s_{n}=x_{1}+x_{2}+\cdots+x_{n}.$ We first prove that $$\sum_{k=1}^{\infty}\dfrac{s_{k+1}-s_{k}}{s_{k+1}}$$diverges.
Since $s_{k}\rightarrow+\infty,$ for all positive integer $m$ there is a positive integer $n$ so that $s_{n+1}>2s_{m}.$ Also $\{s_{k}\}$ is increasing, so$$\begin{aligned}
\sum_{k=m}^{n}\dfrac{s_{k+1}-s_{k}}{s_{k+1}}&\geq\sum_{k=m}^{n}\dfrac{s_{k+1}-s_{k}}{s_{n+1}}=\dfrac{s_{n+1}-s_{m}}{s_{n+1}}\\
&>\dfrac{s_{n+1}-\frac{1}{2}s_{n+1}}{s_{n+1}}=\dfrac{1}{2},\\
\end{aligned}$$
i.e. for all positive integer $m$, there is a positive integer $n$ so that $$\sum_{k=m}^{n}\dfrac{s_{k+1}-s_{k}}{s_{k+1}}>\dfrac{1}{2},$$which shows that $\{\displaystyle\sum_{k=1}^n\dfrac{s_{k+1}-s_{k}}{s_{k+1}}\}$ is not Cauchy. Hence$$\sum_{k=1}^{\infty}\dfrac{s_{k+1}-s_{k}}{s_{k+1}}=+\infty.$$
In the other hand we have $s_{k+1}-s_{k}=x_{k+1},$ hence$$\sum_{k=1}^{\infty}\dfrac{x_{k+1}}{s_{k+1}}=\sum_{k=2}^{\infty}\dfrac{x_{k}}{s_{k}}=+\infty.$$Then let $a_{k}=\dfrac{x_{k}}{s_{k}},$ we have $\lim\limits_{k\rightarrow\infty}\dfrac{a_{k}}{x_{k}}=\lim\limits_{k\rightarrow\infty}\dfrac{1}{s_{k}}=0$ and $\sum\limits_{k=2}^{\infty}a_{k}=+\infty.$
