How do you show whether #sum_(n=2)^oo 1/ln^3(n)# converges or diverges?

The Cauchy condensation test states that if #f(n)# is nonnegative and nonincreasing real sequence, then #sum_(n=1)^oof(n)# converges if and only if the sum #sum_(n=0)^oo2^nf(2^n)# converges as well.

Let #f(n) = {(1/ln^3(n) if n>1),(f(n+1) if n=1):}#

As adding a finite value to a series does not change whether it converges or diverges, we can add #1/ln^3(2)# to the given series without changing the result. Doing so, we can apply the Cauchy condensation test:

#1/ln^3(2)+sum_(n=2)^oo1/ln^3(n) = sum_(n=1)^oof(n)# converges if and only if #sum_(n=0)^oo2^nf(2^n) = sum_(n=1)^oo2^n/ln^3(2^n)# converges.

Now, looking at the condensed sum, we have

#sum_(n=1)^oo2^n/ln^3(2^n) = sum_(n=1)^oo2^n/(nln(2))^3=sum_(n=1)^ooln^(-3)(2)2^n/n^3#

which diverges by the divergence test, as #lim_(n->oo)|2^n/n^3|=oo#.

As #sum_(n=0)^oo2^nf(2^n)# diverges, so does #sum_(n=1)^oof(n)=1/ln^3(2)+sum_(n=2)^oo1/ln^3(n)#, and thus so does the given series.

My first intuition is to show that #1/ln^3(n)>1/n# to claim that #sum_(n=2)^oo1/ln^3(n)# diverges through direct comparison, since #sum_(n=2)^oo1/n# diverges as the harmonic series.

Notice that #1/ln^3(n)=1/n(n/ln^3(n))#. From this we see that #1/ln^3(n)>1/n# if #n/ln^3(n)>1#.

To see if this is true as these series extend infinitely, we can take the infinite limit of #n/ln^3(n)#.

#lim_(nrarroo)n/ln^3(n)#

This is indeterminate in the form #oo/oo#, so we can apply L'Hospital's rule and tale the derivative of the numerator and denominator separately.

#=lim_(nrarroo)1/(3ln^2(n)1/n)=lim_(nrarroo)n/(3ln^2(n))#

Reapplying L'Hospitals (we can see a pattern forming):

#=lim_(nrarroo)1/(6ln(n)1/n)=lim_(nrarroo)n/(6ln(n))#

L'Hospital's once more:

#=lim_(nrarroo)1/(6 1/n)=lim_(nrarroo)n/6=oo#

Thus we've seen that #n/ln^3(n)# will be far greater than #1#, and that we can state that #1/ln^3(n)=1/n(n/ln^3(n))>1/n# for sufficiently large values of #n#.

Through direct comparison, #sum_(n=2)^oo1/ln^3(n)# diverges.