Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.frac13+frac15+frac17+frac19+frac{1}{11}+...

naivlingr

naivlingr

Answered question

2021-02-09

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
13+15+17+19+111+...

Answer & Explanation

tabuordy

tabuordy

Skilled2021-02-10Added 90 answers

Given: The series
13+15+17+19+111+...
To determine: The convergence and divergence of the given series by Integral test.
Explanation:
Integral Test: Let f(x) be a positive, continuous and decreasing function for x[1,) and f(n)=an where n is any natural number then the improper integral 1f(x)dx and infinite n=1an both converge or diverge simultaneously.
Here the infinite sum is 13+15+17+19+111+...=n=112n+1
Therefore an=12n+1
So let f(x)=12x+1 for x[1,)
Then clearly f(x) is continuous, positive, and decreasing for x[1,) and f(n)=12n+1=an for any natural number n.
Hence Integral Test can be applied for f(x)=12x+1 and an12n+1
Now checking for the convergence or divergence of the improper integral
1f(x)dx for f(x)=12x+1
So,
1f(x)dx
=limt1t12x+1dx
=12limt1t22x+1dx [multiplying and dividing by 2]
=12limt[ln(2x+1)]1t[f(x)f(x)dx=ln[f(x)]]
=12limt[ln(2t+1)ln3]
=
Hence 1f(x)dx=112x+1dx diverges
Therefore by Integral test
n=112n+1=13+15+17+19+111+... also diverge.
Answer: The series 13+15+17+19+111+... is divergent

Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-26Added 2605 answers

Answer is given below (on video)

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