Use the limit Comparison Test to detemine the convergence or divergence of the series. sum_{n=1}^inftyfrac{5}{n+sqrt{n^2+4}}

CoormaBak9

CoormaBak9

Answered question

2020-11-24

Use the limit Comparison Test to detemine the convergence or divergence of the series.
n=15n+n2+4

Answer & Explanation

Arnold Odonnell

Arnold Odonnell

Skilled2020-11-25Added 109 answers

Given Data
The series is n=15n+n2+4
Let,
an=5n+n2+4
bn=1n
Evaluate the convergence or divergence of the series by using limit ratio test,
limnanbn=limn(5n+n2+4)(1n)
=limn(5nn+n2+4)
=limn(5nn+n1+4n2)
=limn(5nn(1+1+4n2))
=51+1+4
=51+1
=52
>0
The value of series by using the limit comparison test is 52, which is greater than 0.
Hence the series is divergent.

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