Constructing a divergent series Please would you help me with this question? I've been thinking about it for ages but I've made very little headway, so if possible a hint would be ideal. 
Let  $\sum_{n=1}^∞{x_n}$ be a divergent series, where $x_n > 0$ for all $n$. Show that there is a divergent series  $\sum_{n=1}^∞{y_n}$ with $y_n > 0$ for all $n$, such that $(\frac{y_n}{x_n}) → 0.$
I have not been taught analysis formally, hence my lack of progress. I know to consider the series as a sequence of partial sums, and I tried to take the contrapositive of the statement but that just overcomplicated matters. I know I don't have many ideas to present but I have been trying this for days.
Thank you in advance. 
 A: You can break up your divergent series $\sum x_n$ into bunches of
consecutive terms, each of which are $>1$. In detail there are
$n_1$, $n_2,\ldots$ with
$$x_1+x_2+\cdots +x_{n_1}>1,$$
$$x_{n_1+1}+x_{n_1+2}+\cdots +x_{n_2}>1,$$
etc. Pick suitable constants $u_1,u_2,\ldots$ and set $y_k=u_ix_k$
whenever $n_{i-1}<k\le n_i$.
A: There are some 'clever' ways to do this, but my favourite way falls into the 'just do it' category.  
We look for a sequence of the form
$$
y_n=a_nx_n
$$
where $a_n\to 0$.  To simplify things, let's let $(a_n)$ be of the form:
$$
\underbrace{1,\cdots,1}_{N_1},\underbrace{\frac{1}{2},\cdots,\frac12}_{N_2},\underbrace{\frac13,\cdots,\frac13}_{N_3},\underbrace{\frac14,\cdots\frac14}_{N_4}\cdots
$$
for suitable constants $N_1,N_2,\cdots$.  
Now, since we know that $\sum_{n=1}^\infty x_n$ is divergent, we can pick $N_1,N_2,\cdots$ to be such that
\begin{gather}
x_1+\cdots + x_{N_1} \ge 1 \\
x_{N_1 + 1} + \cdots + x_{N_2} \ge 2 \\
x_{N_2 + 1} + \cdots + x_{N_3} \ge 3
\end{gather}
and so on.  With such a choice of $N_1,N_2,\cdots$, we see that\begin{gather}
y_1+\cdots + y_{N_1} \ge 1 \\
y_{N_1 + 1} + \cdots + y_{N_2} \ge 1 \\
y_{N_2 + 1} + \cdots + y_{N_3} \ge 1
\end{gather}
and so $\sum_{n=1}^\infty y_n$ is divergent.
A: Based on this result:
If the positive series $\sum a_n$ diverges and $s_n=\sum\limits_{k\leqslant n}a_k$ then $\sum \frac{a_n}{s_n}$ diverges as well
Let's take $S_n=\sum\limits_{k=1}^n x_n\to+\infty\quad$ then $\quad\displaystyle y_n=\frac{x_n}{S_n}$ agrees with your requirements.


*

*$x_n>0\implies S_n>0\implies y_n>0$

*$\displaystyle \frac{y_n}{x_n}=\frac 1{S_n}\to 0$

*$\sum\ y_n$ diverges

For instance the classical divergent series $x_n=\frac 1n\ $ gives $\  y_n\sim\frac 1{n\ln(n)}$ 
A: To make it even easier:
Define $(y_{n})$ as $y_{1}=\frac{x_{1}}{1}, y_{2}=y_{3}=\frac{x_{2}}{2}, y_{4}=y_{5}=y_{6}=\frac{x_{3}}{3},y_{7}=y_{8}=y_{9}=y_{10}=\frac{x_{4}}{4}\dots$ where every time $\frac{x_{n}}{n}$ $n$ is repeated $n$ times...
