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 2021-02-09 Answered

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+...

You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

tabuordy
Answered 2021-02-10 Author has 91 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

Not exactly what you’re looking for?
Ask My Question
Jeffrey Jordon
Answered 2021-12-26 Author has 2313 answers

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-01-24

Reaching upon 9=1 while solving x for 3tan(x15)=tan(x+15)
Substituting y=x+45, we get
3tan(y60)=tan(y30)
3tany31+3tany=tany131+13tany

3(tan2y3)=3tan21

9=1

The solution provided by the book x=nπ+π4 fits, so why did i get 9=1?

asked 2020-12-15
In the following items, you will analyze how several transformations affect the graph of the function f(x)=1x. Investigate the graphs of f(x)=1x,g(x)=f(x)=1x+2,h(x)=1x2,p(x)=1x4 and z(x)=1x2+1. If you use a graphing calculator, select a viewing window ±23.5 for x and ±15.5 for y. At what values in the domain did vertical asymptotes occur for each of the functions? Explain why the vertical asymptotes occur at these values.
asked 2022-06-21
Find the value of cos tan 1 sin cot 1 ( x )
... Let cot 1 x = z
x = cot z
Then,
sin cot 1 ( x )
= sin z
= 1 csc z
= 1 1 + cot 2 z
= 1 1 + x 2
asked 2022-04-16
How do I rewrite ((cost)3)2((cost)((sint)2)) to 3(cost)32cost ?
asked 2022-06-13
Verifying the identity 1 cos x cos x = sin x tan x
I have tried a few things and nothing works
Left Side
1 cos x cos x
1 ( cos x ) 2
1 1 + cos 2 x 2
cos 2 x 2
cos x 2 × cos 2
... And I am out in left field!
asked 2020-11-08
If A=[120110140],B=[123111222] and C=[123111111]. Show that AB=AC but BC
asked 2022-04-30
Prove that
(sin2x2)4331sinx+1x23}<1

New questions