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
Answer:
It is going to be G - 3,024 m²
Step-by-step explanation:
:)
Answer:
-0.38
Step-by-step explanation:
0.6-(0.7)(1.4)
0.6-0.98
-0.38
The problem would be written out like this: n/5 + 4 = 16. We would have to subtract 4 from both sides to get n/5 = 12. Then, we multiply both sides by 5 and after that step, you get 60. n = 60.
