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

1. Diverges

2. Converges

3. Diverges

Step-by-step explanation:

Solution:-

Limit comparison test:

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

                          Lim ( n-> ∞ )    = c

Where,

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

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

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

a) The given series ∑ is:

                   (n = 1) ∑^∞  

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

                  (n = 1) ∑^∞  

- Apply the limit ( n - > ∞ ):

                 (n = 1) ∑^∞      .... The comparative series ( ∑ )

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

- Compute the limit:

                 Lim ( n-> ∞ )  

                 Lim ( n-> ∞ )    

- Apply the limit ( n - > ∞ ):

                Lim ( n-> ∞ )    = = 1   ... Finite

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

- We check for ∑ convergence or divergence.

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

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

Diverges    

b)          

The given series ∑ is:

                   (n = 1) ∑^∞  

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

                  (n = 1) ∑^∞  

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

                  (n = 1) ∑^∞     .... The comparative series ( ∑ )

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

- Compute the limit:

                 Lim ( n-> ∞ )  

                 Lim ( n-> ∞ )    

- Apply the limit ( n - > ∞ ):

                Lim ( n-> ∞ )    = = 1   ... Finite

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

- We check for ∑ convergence or divergence.

- The ∑ = ( ) converges by p-series test ∑ ( ) where p = 3/2 > 1. Hence, ∑

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

converges

Comparison Test:-  

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

-Then the following conditions are applied:

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

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

c) The given series ∑ is:

                   (n = 1) ∑^∞

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

                  (n = 1) ∑^∞

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

                 (n = 1) ∑^∞   ... The comparative series ( ∑ )

- Compute the difference between sequences ( - ):

                 , for all values of ( n )

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

               ∑ = (n = 1) ∑^∞  diverges

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

Diverges