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I want to use the integral test to show that $ \sum_{n=3}^{\infty} {1 \over {n\cdot \log{n} \cdot \log{\log {n}}}} $ diverges.

First, I let $ f(x) = {1 \over {x\cdot \log{x} \cdot \log{\log {x}}}} $

I have learned that in order to use the Integral test, $f(x)$ must be continuous, positive, and decreasing at the interval $ [1,\infty) $

However, when I drew the graph of $f(x)$ using an online graphing calculator, the graph only seemed to be satisfying the three conditions of the Integral test when $ x \geq 10 $, or the interval $ [10, \infty) $

Doesn't that mean that I cannot use the Integral test? Other than that, I was also confused on why the series starts from n = 3 when the series is also defined a value at n = 2. (although negative, like at n = 3)

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    $\begingroup$ The starting point doesn't matter. Here $\log\log 2<0$ so it's good to avoid $n=2$. Also $f$ is decreasing for $x>e$, no matter what your "online graphing calculator" says. $\endgroup$ – Lord Shark the Unknown Jul 17 at 7:23
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    $\begingroup$ The integral test works for any function $f(x)$ that is merely eventually continuous, positive, and decreasing. $\endgroup$ – Greg Martin Jul 17 at 7:24
  • $\begingroup$ @GregMartin Okay so does that mean if $f(x)$ satisfies the condition in the interval $[a,\infty)$, I integrate $f(x)$ from $a$ to $\infty$? $\endgroup$ – linearAlg Jul 17 at 7:26
  • $\begingroup$ @LordSharktheUnknown I used Desmos website as my online graphing calculator. Is the website unreliable? $\endgroup$ – linearAlg Jul 17 at 7:26
  • $\begingroup$ @LordSharktheUnknown Never mind I just realized that Desmos uses ln as its natural logarithm. Sorry for bothering you with my silly mistake $\endgroup$ – linearAlg Jul 17 at 7:27
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Since $(\ln(f(x)))' =\dfrac{f'(x)}{f(x)} $,

$\begin{array}\\ (\ln(\ln(\ln(x))))' &=\dfrac{(\ln(\ln(x)))'}{\ln(\ln(x))}\\ &=\dfrac{\dfrac{(\ln(x))'}{\ln(x)}}{\ln(\ln(x))}\\ &=\dfrac{\dfrac{(\ln(x))'}{\ln(x)}}{\ln(\ln(x))}\\ &=\dfrac{1}{x\ln(x)\ln(\ln(x))}\\ \end{array} $

so

$\int \dfrac{dx}{x\ln(x)\ln(\ln(x))} =\ln(\ln(\ln(x))) $ and this goes to $\infty$ as $x \to \infty$.

By looking at $\ln(\ln(...(\ln(x))...) $ nested $m$ deep, we get $(\ln(\ln(...(\ln(x))...))' =\dfrac1{x\ln(x)\ln(\ln(x))...\ln(\ln(...(\ln(x))...)} $ so the integral of that diverges, though extremely slowly.

To formalize this, let $L_0(x) = x$ and $L_{m+1}(x) = \ln(L_{m}(x)) $ for $m \ge 0$.

Then $(L_{m+1}(x))' =(\ln(L_m(x)))' =\dfrac{(L_m(x))'}{L_m(x)} $.

If $(L_m(x))' =\dfrac1{\prod_{k=0}^{m-1} L_k(x)} $, which is true for $m=0$ and $m=1$, then $(L_{m+1}(x))' =\dfrac{(L_m(x))'}{L_m(x)} =\dfrac1{L_m(x)\prod_{k=0}^{m-1} L_k(x)} =\dfrac1{\prod_{k=0}^{m} L_k(x)} $ for all $m \ge 0$.

Therefore $\int \dfrac{dx}{\prod_{k=0}^{m} L_k(x)} =L_{m+1}(x) $ and since the right side diverges, so does the left as $x \to \infty$.

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