Showing $\lim_{x \rightarrow \infty}\int_{4}^{x}\frac{1}{x\log(x)\log \log(x)}$ converges or divegences In the text "Elementary Theory of Calculus" I'm having trouble applying the integral test to show that the series in $(1.)$ converges.
$(1.)$
$$\sum_{}^{}\frac{1}{n\log(n)\log \log(n)}$$
My attack on $(1.)$ can be seen within $(2.)$ yielded the following observations:
$(2.)$
$$\sum_{}^{}\frac{1}{n\log(n)\log \log(n)} <=> \int_{4}^{\infty}\frac{1}{x\log(x)\log \log(x)}$$
$$\lim_{x \rightarrow \infty}\int_{4}^{x}\frac{1}{x\log(x)\log \log(x)} <=> \int_{4}^{x}(\lim_{x \rightarrow \infty}\frac{1}{x\log(x)\log \log(x)})$$
On the RHS side $(2.)$ is appropriate to pass the limit inside the integral or is their a better way to approach proving the integral converges or diverges ?
 A: We have
$$ \frac{d}{dx} \log\log\log{x} = \frac{1}{x\log{x}\log{\log{x}}}. $$
It should be fairly clear what happens to this antiderivative for large $x$.
And no, you certainly can't pass the limit inside the integral:
$$ \int_4^X \frac{dx}{x\log{x}\log{\log{x}}} $$
is a function of one variable: $X$. This is but one reason that it's so important to use different letters for the integration variables and the limits. See also this post.
A: There seem to be a few misunderstandings here. First, the integral test tells you that for a decreasing function $f(x)$ such that $\lim_{x\to\infty} f(x) = 0$, we have that $$\int^\infty_k f(x) dx \,\,\,\,\,\, \text{ and } \,\,\,\,\,\, \sum^\infty_{n=k} f(n)$$ either both converge or both diverge. It does not say that the two are equal. Secondly, $$\lim_{x\to \infty} \int^x_{4} f(x) dx \neq \int^x_4 \left[\lim_{x\to \infty} f(x) \right]dx.$$ Instead, you need to find an anti-derivative $F$ of $f$ and then $$\lim_{x\to\infty}\int^x_{4} f(x) dx = \lim_{x\to\infty} (F(x) - F(4));$$ i.e., you need to take the anti-derivative before considering the limit. 
In your case, note that you can make the substitution $u = \log\log x$ and see that \begin{align*} \int_4^\infty \frac{dx}{x \log x \log \log x} &= \int^\infty_{\log\log4} \frac{du}{u} \\&= \log(u)\bigg|^\infty_{\log\log 4} \\&=  \log\log\log x\bigg|^\infty_{4}\\&= \lim_{t\to\infty} \log\log\log(t) - \log\log\log(4) = \infty.\end{align*} Thus the integral diverges and so does the sum.
