How do you test the alternating series $Sigma (-1)^n/(ln(lnn))$ from n is $[3,oo)$ for convergence?

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Answer 1 · 2018-06-08T23:33:21

The series:

$sum_(n=3)^oo (-1)^n/ln(lnn)$

is convergent.

We have that:

$lim_(n->oo) 1/ ln(ln(n)) = 0$

Consider the function:

$f(x) = 1/ln(ln(x))$

As:

$f'(x) = -1/(ln(lnx))^2 1/lnx 1/x < 0$ for $x > 1$

the function is strictly decreasing in $[1,+oo)$ and thus:

$1/ln(ln(n+1)) < 1/ln(ln(n))$

thus the series:

$sum_(n=3)^oo (-1)^n/ln(lnn)$

is convergent based on Leibniz' theorem.

How do you test the alternating series Sigma (-1)^n/(ln(lnn))Sigma (-1)^n/(ln(lnn)) from n is [3,oo)[3,oo) for convergence?