How do you test the series #Sigma n/sqrt(n^3+1)# from n is #[0,oo)# for convergence?

Question

26753 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T12:17:35

The series diverges.

I would limit compare to #sum1/sqrt(n)#.

I come up with this by looking at dominant terms in the numerator and denominator of the nth term of the given series:

numerator: dominant term is #n#
denominator: dominant term is #sqrt(n^3) = n^(3/2)#

So the given series can be compared to #sum(n/n^(3/2))=sum(1/n^(1/2))=sum(1/sqrt(n))#.

We know that #sum1/sqrt(n)# is a divergent #p#-series with #p=1/2# which is less than 1.

So we calculate #lim_(n to oo)((n/sqrt(n^3+1))/(1/sqrt(n))) #

#=lim_(n to oo)(n/sqrt(n^3+1)*sqrt(n)) =lim_(n to oo)(n^(3/2)/sqrt(n^3+1))#

#=lim_(n to oo)(sqrt(n^(3)/(n^3+1))) = 1#

Since the limit is positive and finite and #sum1/sqrt(n)# diverges, so does #sum(n/sqrt(n^3+1))# by the limit comparison test.