Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑n=1∞2n2−13n5+2n+1

Kyran Hudson

Kyran Hudson

Answered

2021-03-08

Use the Limit Comparison Test to determine the convergence or divergence of the series.
n=12n213n5+2n+1

Answer & Explanation

irwchh

irwchh

Expert

2021-03-09Added 102 answers

We have to check the given series
n=12n213n5+2n+1 is convergent or divergent using the limit comparision test.
According to limit comparision test if two series n=1an and n=1bn with an>0,bn>0 for all n.
Then if limnanbn=c with 0 then either both series converges or both series divergent.
Let n=1an=n=1an2n213n5+2n+1
Take common highest power n from denominator and numerator we get
n=12n213n5+2n+1=n=1n2(21n2)n5(3+2n4+1n5)
=n=1(21n2)n3(3+2n4+1n5)
So, n=1an=n=1(21n2)n3(3+2n4+1n5)
Let another series bn=1n3
bn is convergent p-series since p=3
Now,
limnanbn=limn(21n2)n3(3+2n4+1n5)1n3
=limn(21n2)(3+2n4+1n5)
=203+0+0
=23
=limnanbn=23
which is finite and positive.
Therefore we can conclude by limit comparison test series n=1an will be convergent.
So given series is convergent.

Jeffrey Jordon

Jeffrey Jordon

Expert

2021-12-27Added 2575 answers

Answer is given below (on video)

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