Mathematics, 12.06.2020 02:57 3peak101

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

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Answer from: gretchcampbell

1. Diverges

2. Converges

3. Diverges

Step-by-step 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-> ∞ ) $[ \frac{a_n}{b_n} ]$ = 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) ∑^∞ $[ \frac{n^2+1}{2n^3-1} ]$

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

(n = 1) ∑^∞ $[ \frac{n^2( 1 + \frac{1}{n^2} )}{n^3 ( 2 - \frac{1}{n^2} ) } ] = [ \frac{( 1 + \frac{1}{n^2} )}{n( 2 - \frac{1}{n^2} ) } ]$

- Apply the limit ( n - > ∞ ):

(n = 1) ∑^∞ $[ \frac{1}{2n}]$ .... 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-> ∞ ) $[ \frac{n^2 + 1}{2n^3 - 1} * 2n ] = [ \frac{2n^3 + 2n}{2n^3 - 1} ]$

Lim ( n-> ∞ ) $[ \frac{2n^3 ( 1 + \frac{1}{n^2} ) }{2n^3 ( 1 - \frac{1}{2n^3} ) } ] = [ \frac{ 1 + \frac{1}{n^2} }{ 1 - \frac{1}{2n^3} } ]$

- Apply the limit ( n - > ∞ ):

Lim ( n-> ∞ ) $[ \frac{a_n}{b_n} ]$ = $[ \frac{1 + 0}{1 + 0} ]$ = 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 ∑ ( $\frac{1}{n^p}$ ) where p = 1 ≤ 1. Hence, ∑

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

Diverges

The given series ∑ a_n is:

(n = 1) ∑^∞ $[ \frac{n}{n^\frac{5}{2} +5} ]$

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

(n = 1) ∑^∞ $[ \frac{n( 1 )}{n ( n^\frac{3}{2} + \frac{5}{n} ) } ] = [\frac{1}{( n^\frac{3}{2} + \frac{5}{n} )} ]$

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

(n = 1) ∑^∞ $[ \frac{1}{n^\frac{3}{2} } ]$ .... 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-> ∞ ) $[ \frac{n}{n^\frac{5}{2} +5} * n^\frac{3}{2} ] = [ \frac{n^\frac{5}{2}}{n^\frac{5}{2} +5} ]$

Lim ( n-> ∞ ) $[ \frac{n^\frac{5}{2}}{n^\frac{5}{2} ( 1 + \frac{5}{n^\frac{5}{2}}) } ] = [ \frac{1}{1 + \frac{5}{n^\frac{5}{2}} } ]$

- Apply the limit ( n - > ∞ ):

Lim ( n-> ∞ ) $[ \frac{a_n}{b_n} ]$ = $[\frac{1}{1 + 0}]$ = 1 ... Finite

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

- We check for ∑ b_n convergence or divergence.

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

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

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) ∑^∞ $[ \frac{4 + 3^2}{2^n} ]$

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

(n = 1) ∑^∞ $[ \frac{3^n ( \frac{4}{3^n} + 1 )}{2^n} ]$

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

(n = 1) ∑^∞ $[ \frac{3^n}{2^n} ]$ ... The comparative series ( ∑ b_n )

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

$a_n - b_n = \frac{4 + 3^n}{2^n} - [ \frac{3^n}{2^n} ] \\\\a_n - b_n = \frac{4 }{2^n} \geq 0$ , for all values of ( n )

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

∑ b_n = (n = 1) ∑^∞ $[ \frac{3^n}{2^n} ]$ diverges

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

Diverges

Answer from: Quest

( ͡° ͜ʖ ͡°)

step-by-step explanation:

just copy and paste it.

visit this site for a tutorial ʖ-tutorial-flash-cards/

Answer from: Quest

answer: 38.47 is your answer

step-by-step explanation: hope i you and who ever sees this answer have a great rest of your day : )

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