Show that $x_n=\sum_{k=3}^n \frac{1}{k\ln^4(\ln k)}$ is Cauchy sequence.

Question

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?

Answer 1

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 $K=C_{1/8}^{-4}$.

Answer 2

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?