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)
 A: 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 $x \to \infty$.
