Respuesta :
Answer:
1. Diverges
2. Converges
3. Diverges
Step-by-step explanation:
Solution:-
Limit comparison test:
- Given, ∑[tex]a_n[/tex] and suppose ∑[tex]b_n[/tex] such that both series are positive for all values of ( n ). Then the following three conditions are applicable for the limit:
Lim ( n-> ∞ ) [tex][ \frac{a_n}{b_n} ][/tex] = c
Where,
1) If c is finite: 0 < c < 1, then both series ∑[tex]a_n[/tex] and ∑[tex]b_n[/tex] either converges or diverges.
2) If c = 0, then ∑[tex]a_n[/tex] converges only if ∑[tex]b_n[/tex] converges.
3) If c = ∞ or undefined, then ∑[tex]a_n[/tex] diverges only if ∑[tex]b_n[/tex] diverges.
a) The given series ∑[tex]a_n[/tex] is:
(n = 1) ∑^∞ [tex][ \frac{n^2+1}{2n^3-1} ][/tex]
- We will make an educated guess on the comparative series ∑[tex]b_n[/tex] by the following procedure.
(n = 1) ∑^∞ [tex][ \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} ) } ][/tex]
- Apply the limit ( n - > ∞ ):
(n = 1) ∑^∞ [tex][ \frac{1}{2n}][/tex] .... The comparative series ( ∑[tex]b_n[/tex] )
- Both series ∑[tex]a_n[/tex] and ∑[tex]b_n[/tex] are positive series. You can check by plugging various real number for ( n ) in both series.
- Compute the limit:
Lim ( n-> ∞ ) [tex][ \frac{n^2 + 1}{2n^3 - 1} * 2n ] = [ \frac{2n^3 + 2n}{2n^3 - 1} ][/tex]
Lim ( n-> ∞ ) [tex][ \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} } ][/tex]
- Apply the limit ( n - > ∞ ):
Lim ( n-> ∞ ) [tex][ \frac{a_n}{b_n} ][/tex] = [tex][ \frac{1 + 0}{1 + 0} ][/tex] = 1 ... Finite
- So from first condition both series either converge or diverge.
- We check for ∑[tex]b_n[/tex] convergence or divergence.
- The ∑[tex]b_n[/tex] = ( 1 / 2n ) resembles harmonic series ∑ ( 1 / n ) which diverges by p-series test ∑ ( [tex]\frac{1}{n^p}[/tex] ) where p = 1 ≤ 1. Hence, ∑
- In combination of limit test and the divergence of ∑[tex]b_n[/tex], the series ∑[tex]a_n[/tex] given also diverges.
Answer: Diverges
b)
The given series ∑[tex]a_n[/tex] is:
(n = 1) ∑^∞ [tex][ \frac{n}{n^\frac{5}{2} +5} ][/tex]
- We will make an educated guess on the comparative series ∑[tex]b_n[/tex] by the following procedure.
(n = 1) ∑^∞ [tex][ \frac{n( 1 )}{n ( n^\frac{3}{2} + \frac{5}{n} ) } ] = [\frac{1}{( n^\frac{3}{2} + \frac{5}{n} )} ][/tex]
- Apply the limit ( n - > ∞ ) in the denominator for ( 5 / n ), only the dominant term n^(3/2) is left:
(n = 1) ∑^∞ [tex][ \frac{1}{n^\frac{3}{2} } ][/tex] .... The comparative series ( ∑[tex]b_n[/tex] )
- Both series ∑[tex]a_n[/tex] and ∑[tex]b_n[/tex] are positive series. You can check by plugging various real number for ( n ) in both series.
- Compute the limit:
Lim ( n-> ∞ ) [tex][ \frac{n}{n^\frac{5}{2} +5} * n^\frac{3}{2} ] = [ \frac{n^\frac{5}{2}}{n^\frac{5}{2} +5} ][/tex]
Lim ( n-> ∞ ) [tex][ \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}} } ][/tex]
- Apply the limit ( n - > ∞ ):
Lim ( n-> ∞ ) [tex][ \frac{a_n}{b_n} ][/tex] = [tex][\frac{1}{1 + 0}][/tex] = 1 ... Finite
- So from first condition both series either converge or diverge.
- We check for ∑[tex]b_n[/tex] convergence or divergence.
- The ∑[tex]b_n[/tex] = ( [tex][ \frac{1}{n^\frac{3}{2} } ][/tex] ) converges by p-series test ∑ ( [tex]\frac{1}{n^p}[/tex] ) where p = 3/2 > 1. Hence, ∑
- In combination of limit test and the divergence of ∑[tex]b_n[/tex], the series ∑[tex]a_n[/tex] given also converges.
Answer: converges
Comparison Test:-
- Given, ∑[tex]a_n[/tex] and suppose ∑[tex]b_n[/tex] such that both series are positive for all values of ( n ).
-Then the following conditions are applied:
1 ) If ( [tex]a_n[/tex] - [tex]b_n[/tex] ) < 0 , then ∑[tex]a_n[/tex] diverges only if ∑[tex]b_n[/tex] diverges
2 ) If ( [tex]a_n[/tex] - [tex]b_n[/tex] ) ≤ 0 , then ∑[tex]a_n[/tex] converges only if ∑[tex]b_n[/tex] converges
c) The given series ∑[tex]a_n[/tex] is:
(n = 1) ∑^∞ [tex][ \frac{4 + 3^2}{2^n} ][/tex]
- We will make an educated guess on the comparative series ∑[tex]b_n[/tex] by the following procedure.
(n = 1) ∑^∞ [tex][ \frac{3^n ( \frac{4}{3^n} + 1 )}{2^n} ][/tex]
- Apply the limit ( n - > ∞ ) in the numerator for ( 4 / 3^n ), only the dominant terms ( 3^n ) and ( 2^n ) are left:
(n = 1) ∑^∞ [tex][ \frac{3^n}{2^n} ][/tex] ... The comparative series ( ∑[tex]b_n[/tex] )
- Compute the difference between sequences ( [tex]a_n[/tex] - [tex]b_n[/tex] ):
[tex]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[/tex], for all values of ( n )
- Check for divergence of the comparative series ( ∑[tex]b_n[/tex] ), using divergence test:
∑[tex]b_n[/tex] = (n = 1) ∑^∞ [tex][ \frac{3^n}{2^n} ][/tex] diverges
- The first condition is applied when ( [tex]a_n[/tex] - [tex]b_n[/tex] ) ≥ 0, then ∑diverges only if ∑[tex]b_n[/tex] diverges.
Answer: Diverges
