Does the series converge or diverge? If it converges, what is the sum? Show your work.
(∑∞n=1) −4(−1/2)n−1
We can rewrite the series as:
(-4) ∑∞n=1 (-1/2)n-1
Note that the sum inside the parentheses is a geometric series with first term 1 and common ratio -1/2, so we can use the formula for the sum of a geometric series:
∑∞n=1 (-1/2)n-1 = 1/(1-(-1/2)) = 2
Therefore, the original series converges and its sum is:
(-4)(2) = -8
To determine whether the series (∑∞n=1) -4(-1/2)^(n-1) converges or diverges, let's write out the first few terms:
n = 1: -4(-1/2)^(1-1) = -4(-1/2)^0 = -4(1) = -4
n = 2: -4(-1/2)^(2-1) = -4(-1/2)^1 = 4(-1/2) = -2
n = 3: -4(-1/2)^(3-1) = -4(-1/2)^2 = -4(1/4) = -1
n = 4: -4(-1/2)^(4-1) = -4(-1/2)^3 = 4(-1/8) = -1/2
...
From these terms, we can see that the series alternates between -4 and 2, which suggests that it may not converge.
To confirm this, let's look at the absolute values of the terms:
|n = 1: -4(-1/2)^(1-1)| = |-4(-1/2)^0| = |-4(1)| = 4
|n = 2: -4(-1/2)^(2-1)| = |-4(-1/2)^1| = |4(-1/2)| = 2
|n = 3: -4(-1/2)^(3-1)| = |-4(-1/2)^2| = |-4(1/4)| = 1
|n = 4: -4(-1/2)^(4-1)| = |-4(-1/2)^3| = |4(-1/8)| = 1/2
...
The absolute values of the terms are decreasing but not approaching zero. Therefore, the series does not pass the test for convergence known as the Divergence Test.
In conclusion, the series (∑∞n=1) -4(-1/2)^(n-1) diverges.