Sorry if this is too simple of a question.
Let $S_n = \sum \limits _{k=0} ^{n} \dfrac{1}{k}$. For another sequence $\{x_k\}$ define $S'_n = \sum \limits _{k=0} ^{n} x_k$. Is it possible that $S'_n$ diverges and $ lim _{n \rightarrow \infty} \dfrac{S'_n}{S_n} = 0$?
There actually is an interesting exercise in Baby Rudin asking to prove that if $\{a_n\}_{n\geq 1}$ is a (decreasing) sequence of positive real numbers such that $\lim_{n\to +\infty}a_n = 0$ and $\sum_{n\geq 1}a_n = +\infty$, there exists a (decreasing) sequence $\{b_n\}_{n\geq 1}$ of positive real numbers such that $\lim_{n\to +\infty} b_n = 0$, $\sum_{n\geq 1}b_n=+\infty$ and $\lim_{n\to +\infty}\frac{b_n}{a_n}=0$. Here it is a sketch of the proof:
["Deceleration" of divergent series] Let $A_n=a_1+a_2+\ldots+a_n$. $\{A_n\}_{n\geq 1}$ is an increasing sequence of positive numbers such that $A_n\to +\infty$ as $n\to +\infty$. By summation by parts
$$\begin{eqnarray*} \sum_{n=1}^{N}\frac{a_n}{\sqrt{A_n}}&=&\sqrt{A_N}-\sum_{n=1}^{N-1}A_n\left(\frac{\sqrt{A_n}-\sqrt{A_{n+1}}}{\sqrt{A_n A_{n+1}}}\right)\\&=&\sqrt{A_N}+\sum_{n=1}^{N-1}A_n\left(\frac{a_{n+1}}{\sqrt{A_n A_{n+1}}(\sqrt{A_n}+\sqrt{A_{n+1}})}\right)\\&\geq&\sqrt{A_N}+\sum_{n=1}^{N-1}\sqrt{\frac{A_n}{A_{n+1}}}\cdot\frac{a_{n+1}}{2\sqrt{A_{n+1}}}\end{eqnarray*} $$ hence $\sum_{n=1}^{N}\frac{a_n}{\sqrt{A_n}}\sim 2\sqrt{A_N}$ by Cesàro-Stolz and by taking $b_n=\frac{a_n}{\sqrt{A_n}}$ we are done.
More simply, you may consider a sequence of positive numbers $\{c_n\}_{n\geq 1}$ which is divergent to $+\infty$ as slowly as you like (for instance $c_n=\log\log\ldots\log(n+K)$), then define $a_n$ as $c_{n+1}-c_n$. By telescoping $A_n$ diverges exactly like $c_n$.