Answer:
Step-by-step explanation:
Answer:
The series is absolutely convergent.
Step-by-step explanation:
By ratio test, we find the limit as n approaches infinity of
|[a_(n+1)]/a_n|
a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)
a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)
[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]
= |-3n³/2(n+1)³|
= 3n³/2(n+1)³
= (3/2)[1/(1 + 1/n)³]
Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity
= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity
= 3/2 × 1
= 3/2
The series is therefore, absolutely convergent, and the limit is 3/2
14612 is the answer to your math problem. This is so easy. just use a calculator.
Answer:
1.66666667 <-- repeatedly is your answer
Step-by-step explanation:
