Question

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

Answer 1

Answer:

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.

Answer: Diverges    

b)          

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.

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) ∑^∞ $[ \frac{4 + 3^2}{2^n} ]$

- 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.

Answer: Diverges