Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftysinfrac{1}{n}

alesterp

alesterp

Answered question

2021-01-19

Use the Limit Comparison Test to determine the convergence or divergence of the series.
n=1sin1n

Answer & Explanation

2abehn

2abehn

Skilled2021-01-20Added 88 answers

Limit comparison test:
Suppose we have two series an and bn. Define
limnanbn=L
If L is positive and finite then either both series convergent or both series divergent.
Given that,
n=1sin1n
Since we have,
n=1sin1n
=n=1(1n(1n)313!+(1n)515!+...)
[Using sinx=xx33!+x55!+...]
Here
an=(1n(1n)313!+(1n)515!+...)
So
bn=1n
Now evaluate limnanbn, so
limnanbn=limn1n(1n)313!+(1n)515!+...1n
=limn1n[1(1n)213!+(1n)415!+...]1n
=limn[1(1n)213!+(1n)415!+...]
=1
So limnanbn=1>0 which is positive and finite.
Using p series test [1np is convergent if p>1 or divergent if p1], bn is divergent.
Since limnanbn=1>0 is positive and finite. bn is divergent, so using limit comparison test an is also divergent.
Hence n=1sin1n is divergent.
Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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