Trying to review for the exam. Can someone please help me with this series questions. Thank you

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.