Show that $x_n=\sum_{k=3}^n \frac{1}{k\ln^4(\ln k)}$ is Cauchy sequence. I have the following sequence:
$$
x_n = \sum_{k=3}^n \frac{1}{k\ln^4(\ln k)}
$$
And I have to prove that this sequence is Cauchy sequence, i.e $\forall \varepsilon>0 \; \exists N: \forall n \ge N, \; \forall p \in \mathbb N \Rightarrow |x_n - x_{n+p}| < \varepsilon$. So I get:
$$
\frac{1}{(n+1)\ln^4(\ln(n+1))} + \frac{1}{(n+2)\ln^4(\ln(n+2))} + \dots + \frac{1}{(n+p)\ln^4(\ln(n+p))} < \varepsilon
$$
And I stucked with it... But can I use a proof by contradiction? Let
$$
\frac{1}{(n+1)\ln^4(\ln(n+1))} + \frac{1}{(n+2)\ln^4(\ln(n+2))} + \dots + \frac{1}{(n+p)\ln^4(\ln(n+p))} \ge \varepsilon
$$
Then
$$
\lim_{n \to \infty}\left(\frac{1}{(n+1)\ln^4(\ln(n+1))} + \frac{1}{(n+2)\ln^4(\ln(n+2))} + \dots + \frac{1}{(n+p)\ln^4(\ln(n+p))}\right) \ge \varepsilon
$$
So can I process with this limit?
 A: We are going to show that $\{x_n\}$ is unbounded, and as a consequence, that it is not a Cauchy sequence. The function $1/(x\ln^4(\ln x))$ is decreasing. Then
$$
x_n=\sum_{k=3}^n\frac{1}{k\ln^4(\ln k)}\ge\int_3^n\frac{dx}{x\ln^4(\ln x)}=\int_{\ln3}^{\ln n}\frac{dt}{\ln^4t}.
$$
We know that for any $p>0$
$$
\lim_{t\to\infty}\frac{\ln t}{t^p}=0\implies\ln t\le C_p\,t^p
$$
for some constant $C_p>0$ and all $t\ge3$. Then
$$
x_n\ge\frac{1}{C_p^4}\int_{\ln3}^{\ln n}\frac{dt}{t^{4p}}.
$$
Choose $0<p<1/4$, for instance $p=1/8$. Then
$$
x_n\ge K\int_{\ln3}^{\ln n}\frac{dt}{t^{1/2}}=2\,K\bigl(\sqrt{\ln n}-\sqrt{\ln3}\bigr)$$
where $K=C_{1/8}^{-4}$.
A: You want to prove that $|x_n-x_n+p|<\epsilon$, which using your equation becomes
$$
\left|\sum_{k=n}^{k=n+p}\frac{1}{k \ln^4(\ln(k))}\right|<\epsilon \:\forall p
$$
Since $n>3$, $\ln(\ln(n))>0$, thus all terms in the sum are positive so require
$$
\sum_{k=n}^{k=n+p}\frac{1}{k \ln^4(\ln(k))}<\epsilon \:\forall p
$$
Which is what you have. 
Thus if you can show the following:
$$
\exists p \: : \: \sum_{k=n}^{k=n+p}\frac{1}{k \ln^4(\ln(k))}>\epsilon \forall n
$$
you've disproved it. To do a proof by contradiction you'd have to show that this condition was never satisfied, but surely then your just back where you started?
