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.