Use the Limit Comparison Test to prove convergence or divergence of the infinite series. sum_{n=2}^inftyfrac{n}{sqrt{n^3+1}}

UkusakazaL 2021-01-31 Answered
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
n=2nn3+1
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Expert Answer

Arnold Odonnell
Answered 2021-02-01 Author has 109 answers
Let the given expression be represented as,
an=nn3+1
Consider another function bn=1n
According to the limit comparison test-
1. If limn(anbn)= finite positive value, then series nan and nbn behave alike, i.e. either both converge or both diverge.
2. If limn(anbn)=0 and nbn converges, then series nan will also converge.
3. If limn(anbn) diverges, then series nan will also diverge.
Clearly the series n=1bn is a converging series as the term limn1n0
Check fot he series (anbn) as,
limn(anbn)=(nn311n)
=limn(nn31×n)
=limn(n3n31)
=limn(n3n31)
=limn(111n3)
As n, then 1n30,
limn(anbn)=110
=1
Since limn(anbn) is a finite positive value and n=1bn is converging, therefore the limit comparison test implies that, the series n=1an is also converging.
Consider the following,
n=2nn3+1=n=1nn3+1[nn3+1]n=1
=n=1nn3+1113+1
=n=1nn3+112
Consider that the difference between two finite values is also finite.
Thus, it is concluded that the series =n=2nn3+1 is converging.
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Jeffrey Jordon
Answered 2021-12-27 Author has 2262 answers

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