How do you use the limit comparison test on the series #sum_(n=1)^oo(n+1)/(n*sqrt(n))# ?

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Answer 1 · 2018-05-18T14:10:52

The series:

#sum_(n=1)^oo (n+1)/(nsqrtn)#

is divergent.

Note that:

#(n+1)/(nsqrtn) = n/(nsqrtn) +1/(nsqrtn) = 1/sqrtn+1/(nsqrtn)#

as for every #n in NN#:

#1/(nsqrtn) > 0#

it follows that:

#(n+1)/(nsqrtn) > 1/sqrtn#

and as the series:

#sum_(n=1)^oo 1/sqrtn = sum_(n=1)^oo n^(-1/2)#

is divergent based on the #p# series test, we can conclude that:

#sum_(n=1)^oo (n+1)/(nsqrtn)#

is also divergent.