Trying to review for the exam. Can someone please help me with this series questions. Thank you
Use geometric series test to determine whether converges or diverges if it converges find the sum
∞
∑. (π/2)^n
n=1
∞
∑. 3^n/(-2)^n
n=2
π/2 > 1 so it diverges
3^n/(-2)^n = (-1)^n * (3/2)^n and 3/2 > 1 so it diverges.
Recall the error bound on alternating series.
So if it's greater than one it diverges and if its less than one it converges.
To determine whether a series converges or diverges, we can use the geometric series test. The general form of a geometric series is given by:
∑ ar^(n-1) (equation 1)
n=1
where 'a' is the first term, 'r' is the common ratio, and 'n' is the index of summation.
Now let's apply the geometric series test to the given series:
1) ∑ (π/2)^n (n=1, ∞)
The first term 'a' is (π/2)^1 = π/2, and the common ratio 'r' is (π/2)^n / (π/2)^(n-1) = π/2.
For a geometric series to converge, the absolute value of the common ratio 'r' should be less than 1. In this case, |π/2| = π/2, which is greater than 1. Therefore, the series diverges.
2) ∑ 3^n / (-2)^n (n=2, ∞)
To apply the geometric series test, we need to express the series in the form of equation 1:
∑ ar^(n-1)
In this case, the first term 'a' is (3^2) / (-2)^2 = 9/4, and the common ratio 'r' is (3^n / (-2)^n) / (3^(n-1) / (-2)^(n-1)).
Simplifying the expression, we get:
r = (3^n * (-2)^(n-1)) / (3^(n-1) * (-2)^n) = -3/2
Again, for a geometric series to converge, the absolute value of the common ratio 'r' should be less than 1. In this case, |-3/2| = 3/2, which is greater than 1. Therefore, the series diverges.
In conclusion, both series diverge.