Does the series converge or diverge?

Question

Use OCT and #ln(n) < n^k# to see if it converges or diverges.

#sum_(n=2)^oo ln(n)/n^(3/2)#

Instead of using the p-series theorem to solve this, I need to use the natural log theorem.

Theorem: The natural log function ln(x) is eventually below any positive power function #x^k#. This means that ln(n) < n^k for any positive number k, if n is big enough.

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Answer 1 · 2017-07-27T08:31:45

I think you actually need both:

#(1)# we know that #lim_(x->oo) lnx/x^k = 0# for any #k > 0#, then also:

#lim_(n->oo) lnn/n^k = 0#

Choose now #k=1/4# and #epsilon = 1#. By the definition of the limit, we can find #N# such that:

#n>N => lnn/n^(1/4) < 1#

#(2)# For #n > N# we have:

#lnn/n^(3/2) = lnn/n^(1/4)1/n^(5/4) < 1/n^(5/4)#

As #sum 1/n^(5/4)# is a convergent series based on the #p#-series test, by direct comparison also:

#sum_(n=2)^oo lnn/n^(3/2) #

is convergent.