Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{1}{n^2(n^2+4)}

floymdiT 2020-10-23 Answered
Use the Limit Comparison Test to determine the convergence or divergence of the series.
n=11n2(n2+4)
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Expert Answer

okomgcae
Answered 2020-10-24 Author has 93 answers

We have to find the given series is convergence or divergence.
Limit comparison test:
Suppose that we have two series n=1an and n=1bn with an0 and bn>0 for all n
Then, if limnanbn=c, with 0 then either both series convergent or both series divergent.
We have given
n=11n2(n2+4)
Let n=1an=n=11n2(n2+4)
Then, an=1n2(n2+4)
And suppose n=1bn=n=11n4 another series.
Then, bn=1n4
And n=1bn is a p-series and p=4 so, it is convergent.
Now,
limnanbn=limn1a2(a2+4)1a4
=limn1n2(n2+4)×n41
=limnn4n4(1+4a2)
=limn1(1+4a2)
=1 (finite and non zero)
Since, series n=1bn is convergent so, by comparison test series n=1an is convergent.
Therefore, given series is convergent.

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Jeffrey Jordon
Answered 2021-12-27 Author has 2262 answers

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