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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 ?

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    $\begingroup$ You have several very deep errors here: it's not true that the sum and integral are equal to each other (but it is true that they'll converge or diverge together - try and understand why!), and your attempt to interchange limit and order-of-integration makes no sense at all (and above and beyond that, you really shouldn't use the same variable name for your (formal) integration variable and the limit of the integral, because it leads to exactly this sort of confusion.) $\endgroup$ – Steven Stadnicki Jun 5 '17 at 19:29
  • $\begingroup$ Look for how to compute improper integrals. Limit cannot be simply taken inside. $\endgroup$ – Anurag A Jun 5 '17 at 19:30
  • $\begingroup$ i think your integral doesn't converge on the given interval $\endgroup$ – Dr. Sonnhard Graubner Jun 5 '17 at 19:30
  • $\begingroup$ Thanks for the notification The integral of our given function whether it converges or diverges would apply convergence or divergence for our series it was my mistake to think that the sum and integral were equivalent to each other. $\endgroup$ – Zophikel Jun 5 '17 at 19:35
  • $\begingroup$ Let $x=e^{e^u}$ and compare to $\int_1^\infty\frac1x~\mathrm dx$. $\endgroup$ – Simply Beautiful Art Jun 6 '17 at 1:26
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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.

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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.

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