Question

1. Use limit comparison test to determine whether the series converges or diverges:

Σ[infinity]_n=1 n^2 + 1 / 2n^3 - 1

2. Use limit comparison test to determine whether the series converges or diverges:

Σ[infinity]_n = 1 n / √n^5 + 5

3. Use direct comparison test to determine whether the series converges or diverges:

Σ[infinity]_n = 1 4 + 3^n / 2^n

Answer 1

1. Diverges

2. Converges

3. Diverges

Explanation:

Solution:-

Limit comparison test:

- Given, ∑
`a_n` and suppose ∑
`b_n` such that both series are positive for all values of ( n ). Then the following three conditions are applicable for the limit:

Lim ( n-> ∞ )
`[ (a_n)/(b_n) ]` = c

Where,

1) If c is finite: 0 < c < 1, then both series ∑
`a_n` and ∑
`b_n` either converges or diverges.

2) If c = 0, then ∑
`a_n` converges only if ∑
`b_n` converges.

3) If c = ∞ or undefined, then ∑
`a_n` diverges only if ∑
`b_n` diverges.

a) The given series ∑
`a_n` is:

(n = 1) ∑^∞
`[ (n^2+1)/(2n^3-1) ]`

- We will make an educated guess on the comparative series ∑
`b_n` by the following procedure.

(n = 1) ∑^∞
`[ (n^2( 1 + (1)/(n^2) ))/(n^3 ( 2 - (1)/(n^2) ) ) ] = [ (( 1 + (1)/(n^2) ))/(n( 2 - (1)/(n^2) ) ) ]`

- Apply the limit ( n - > ∞ ):

(n = 1) ∑^∞
`[ (1)/(2n)]` .... The comparative series ( ∑
`b_n` )

- Both series ∑
`a_n` and ∑
`b_n` are positive series. You can check by plugging various real number for ( n ) in both series.

- Compute the limit:

Lim ( n-> ∞ )
`[ (n^2 + 1)/(2n^3 - 1) * 2n ] = [ (2n^3 + 2n)/(2n^3 - 1) ]`

Lim ( n-> ∞ )
`[ (2n^3 ( 1 + (1)/(n^2) ) )/(2n^3 ( 1 - (1)/(2n^3) ) ) ] = [ ( 1 + (1)/(n^2) )/( 1 - (1)/(2n^3) ) ]`

- Apply the limit ( n - > ∞ ):

Lim ( n-> ∞ )
`[ (a_n)/(b_n) ]` =
`[ (1 + 0)/(1 + 0) ]` = 1 ... Finite

- So from first condition both series either converge or diverge.

- We check for ∑
`b_n` convergence or divergence.

- The ∑
`b_n` = ( 1 / 2n ) resembles harmonic series ∑ ( 1 / n ) which diverges by p-series test ∑ (
`(1)/(n^p)` ) where p = 1 ≤ 1. Hence, ∑

- In combination of limit test and the divergence of ∑
`b_n`, the series ∑
`a_n` given also diverges.

Answer: Diverges

b)

The given series ∑
`a_n` is:

(n = 1) ∑^∞
`[ (n)/(n^(5)/(2) +5) ]`

- We will make an educated guess on the comparative series ∑
`b_n` by the following procedure.

(n = 1) ∑^∞
`[ (n( 1 ))/(n ( n^(3)/(2) + (5)/(n) ) ) ] = [(1)/(( n^(3)/(2) + (5)/(n) )) ]`

- Apply the limit ( n - > ∞ ) in the denominator for ( 5 / n ), only the dominant term n^(3/2) is left:

(n = 1) ∑^∞
`[ (1)/(n^(3)/(2) ) ]` .... The comparative series ( ∑
`b_n` )

- Both series ∑
`a_n` and ∑
`b_n` are positive series. You can check by plugging various real number for ( n ) in both series.

- Compute the limit:

Lim ( n-> ∞ )
`[ (n)/(n^(5)/(2) +5) * n^(3)/(2) ] = [ (n^(5)/(2))/(n^(5)/(2) +5) ]`

Lim ( n-> ∞ )
`[ (n^(5)/(2))/(n^(5)/(2) ( 1 + (5)/(n^(5)/(2))) ) ] = [ (1)/(1 + (5)/(n^(5)/(2)) ) ]`

- Apply the limit ( n - > ∞ ):

Lim ( n-> ∞ )
`[ (a_n)/(b_n) ]` =
`[(1)/(1 + 0)]` = 1 ... Finite

- So from first condition both series either converge or diverge.

- We check for ∑
`b_n` convergence or divergence.

- The ∑
`b_n` = (
`[ (1)/(n^(3)/(2) ) ]` ) converges by p-series test ∑ (
`(1)/(n^p)` ) where p = 3/2 > 1. Hence, ∑

- In combination of limit test and the divergence of ∑
`b_n`, the series ∑
`a_n` given also converges.

Answer: converges

Comparison Test:-

- Given, ∑
`a_n` and suppose ∑
`b_n` such that both series are positive for all values of ( n ).

-Then the following conditions are applied:

1 ) If (
`a_n` -
`b_n` ) < 0 , then ∑
`a_n` diverges only if ∑
`b_n` diverges

2 ) If (
`a_n` -
`b_n` ) ≤ 0 , then ∑
`a_n` converges only if ∑
`b_n` converges

c) The given series ∑
`a_n` is:

(n = 1) ∑^∞
`[ (4 + 3^2)/(2^n) ]`

- We will make an educated guess on the comparative series ∑
`b_n` by the following procedure.

(n = 1) ∑^∞
`[ (3^n ( (4)/(3^n) + 1 ))/(2^n) ]`

- Apply the limit ( n - > ∞ ) in the numerator for ( 4 / 3^n ), only the dominant terms ( 3^n ) and ( 2^n ) are left:

(n = 1) ∑^∞
`[ (3^n)/(2^n) ]` ... The comparative series ( ∑
`b_n` )

- Compute the difference between sequences (
`a_n` -
`b_n` ):

`a_n - b_n = (4 + 3^n)/(2^n) - [ (3^n)/(2^n) ] \\\\a_n - b_n = (4 )/(2^n) \geq 0`, for all values of ( n )

- Check for divergence of the comparative series ( ∑
`b_n` ), using divergence test:

∑
`b_n` = (n = 1) ∑^∞
`[ (3^n)/(2^n) ]` diverges

- The first condition is applied when (
`a_n` -
`b_n` ) ≥ 0, then ∑diverges only if ∑
`b_n` diverges.

Answer: Diverges