<s>[INST] Cauchy but not rapidly Cauchy I want to show that the sequence $\{\frac{(-1)^n}{n}\}$ is Cauchy but not rapidly Cauchy. Here is the work I done so far. I am curiously if I made any errors.
Consider the normed linear space $\mathbb{R}$ with the norm of the absolute value. Let 
$$
\{f_n\}=\left\{\frac{(-1)^n}{n}\right\}=\left\{-1,\frac{1}{2},\frac{-1}{3},\ldots\right\}.
$$
Suppose $\{f_n\}\rightarrow 0$. Let $\epsilon>0$ and $N\in\mathbb{N}$ such that $n\geq N$. We want to show $|f_n-0|<\epsilon$ in order to prove the convergence of $\{f_n\}\rightarrow 0$. 
Consider
\begin{eqnarray*}
\left|\frac{(-1)^n}{n}-0\right| &<& \epsilon\\
\left|\frac{(-1)^n}{n}\right| &<& \epsilon\\
\left| \frac{1}{n} \right| &<& \epsilon\\
\frac{1}{n}&<& \epsilon
\end{eqnarray*}
Therefore $n>\frac{1}{\epsilon}$. Let $N=\frac{1}{\epsilon}$. Then $n>\frac{1}{\epsilon}$ implies $|f_n-0|<\epsilon$, meaning $\{f_n\}\rightarrow 0$. Since $\{f_n\}$ converges in $\mathbb{R}$,  $\{f_n\}$ is Cauchy.
Now we want to show that $\{f_n\}$ is not rapidly Cauchy. In other words, we want to show that for $\{f_n\}$ there is not a convergent series of positive numbers $\sum_{k=1}^\infty \epsilon_k$ for which
$$
|f_{k+1}-f_k|\leq \epsilon_k^2.
$$
Consider $|f_{k+1}-f_k|\leq \epsilon_k^2$.
\begin{eqnarray*}
|f_{k+1}-f_k| = \left| \frac{(-1)^{k+1}}{k+1}-\frac{(-1)^k}{k} \right| &\leq& \epsilon_k^2\\
\left|\frac{k(-1)^{k+1}-(k+1)(-1)^k}{k(k+1)} \right|&\leq& \epsilon_k^2
\end{eqnarray*}
Suppose $k$ is even. Then
$$
\left|\frac{-k-(k+1)}{k(k+1)} \right| = \left|\frac{-2k-1}{k^2+k} \right| \leq \epsilon_k^2
$$
Suppose $k$ is odd. Then 
$$
\left|\frac{k+(k+1)}{k(k+1)} \right| = \left|\frac{2k+1}{k^2+k} \right| \leq \epsilon_k^2
$$
Therefore
$$
\frac{2k+1}{k^2+k} \leq \epsilon_k^2
$$
So $\epsilon_k=\sqrt{\frac{2k+1}{k^2+k}}$. Consider $\sum\limits_{k=1}^\infty \sqrt{\frac{2k+1}{k^2+k}}$. Notice $\sqrt{\frac{2k}{k^2}}=\frac{\sqrt{2}}{k}\leq\sqrt{\frac{2k+1}{k^2+k}}$ and $\sqrt{2}\sum\limits_{k=1}^\infty\frac{1}{k}$ diverges by the $p$-series since $p=1$. Therefore by the comparison test, $\sum\limits_{k=1}^\infty \sqrt{\frac{2k+1}{k^2+k}}$ diverges. Thus $\{f_n\}$ is not rapidly Cauchy.
 [/INST] Your argument looks correct, although it's a bit long and drawn out. I might argue this way: From your definition we have $f_k$ rapidly Cauchy iff
$$\sum_{k=1}^{\infty} |f_{k+1} - f_k|^{1/2} < \infty.$$
(Never heard of "rapidly Cauchy" before). Now
$$f_{k+1} - f_k = \frac{(-1)^{k+1}}{k+1} - \frac{(-1)^{k} }{k} = (-1)^k\left ( -\frac{1}{k+1} -\frac{1}{k} \right ).$$
So $|f_{k+1} - f_k| = 1/(k+1) + 1/k > 1/k.$ Thus $|f_{k+1} - f_k|^{1/2}> 1/\sqrt k.$ Since $\sum_k 1/\sqrt k = \infty,$ $f_k$ is not rapidly Cauchy
</s>