# Using integral test to show $\sum_{n=3}^{\infty} {1 \over {n\cdot \log{n} \cdot \log{\log {n}}}}$ diverge

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)

• 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. – Lord Shark the Unknown Jul 17 at 7:23
• The integral test works for any function $f(x)$ that is merely eventually continuous, positive, and decreasing. – Greg Martin Jul 17 at 7:24
• @GregMartin Okay so does that mean if $f(x)$ satisfies the condition in the interval $[a,\infty)$, I integrate $f(x)$ from $a$ to $\infty$? – linearAlg Jul 17 at 7:26
• @LordSharktheUnknown I used Desmos website as my online graphing calculator. Is the website unreliable? – linearAlg Jul 17 at 7:26
• @LordSharktheUnknown Never mind I just realized that Desmos uses ln as its natural logarithm. Sorry for bothering you with my silly mistake – linearAlg Jul 17 at 7:27

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 $$x \to \infty$$.