How do you use the integral test to determine if #ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+...# is convergent or divergent?

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mason m Share
Jan 3, 2017

Answer:

It is divergent. See explanation.

Explanation:

First write the series:

#ln2/2+ln3/3+ln4/4+...=sum_(n=2)^ooln(n)/n#

Before getting into the integral test, we must assure two things first: for the integral test to apply to #sum_(n=N)^oof(n)# we must have #a_n>0# for the given interval and #f(n)# must be decreasing on the same interval.

Both of these are true since #ln(n)>0# and #n>0# on #n in [2,oo)#, so #ln(n)/n>0# on the same interval.

Furthermore, #ln(n)# grows slower than #n# so we see that #ln(n)/n# is decreasing because #n# overpowers #ln(n)# in the numerator. You can also show this by taking the derivative of #ln(n)/n# and showing it's always negative on #n in [2,oo)#.

So, we see the integral test applies. The integral test states that if the two aforementioned conditions are met, then for #sum_(n=N)^oof(n)#, evaluate the improper integral #int_N^oof(x)dx#.

If the integral converges to a real, finite value, then the series converges. If the integral diverges, then the series does too.

So, we take the integral #int_2^ooln(x)/xdx#. Take the limit as it goes to infinity:

#int_2^ooln(x)/xdx=lim_(brarroo)int_2^bln(x)/xdx#

Letting #u=ln(x)#, so #du=1/xdx#:

#=lim_(brarroo)int_ln(2)^ln(b)ucolor(white).du#

#=lim_(brarroo)[1/2u^2]_ln(2)^ln(b)#

#=lim_(brarroo)1/2ln^2(b)-1/2ln^2(2)#

As #brarroo#, we see that #ln(b)rarroo#, so:

#=oo#

The integral diverges. Thus, we see that #sum_(n=2)^ooln(n)/n# diverges as well.

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