find the series radius and interval of convergence.

from n=0 to infinity
an=[(-1)^(n+1)(x+2)^n]/(n2^n)

To find the radius and interval of convergence of the given series, we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit equals 1, the ratio test is inconclusive, and more tests are needed.

Let's apply the ratio test to the given series:

Step 1: Take the absolute value of the terms in the series.
|an| = |[(-1)^(n+1)(x+2)^n]/(n2^n)|

Step 2: Find the ratio of consecutive terms.
|(an+1) / an| = |{[(-1)^(n+2)(x+2)^(n+1)]/[(n+1)2^(n+1)]} * {(n2^n) / [(-1)^(n+1)(x+2)^n]}|

Step 3: Simplify the ratio.
|(an+1) / an| = |(-1)^(n+2)(x+2)^(n+1) / (n+1)2^(n+1) * (-1)^(n+1)(x+2)^n / (n2^n)|

Step 4: Cancel out like terms.
|(an+1) / an| = |-1 * (-1) * (x+2)^1 / (n+1) * 1 / (n2)|

Step 5: Take the absolute value of the ratio.
|(an+1) / an| = |(x+2) / [(n+1) * 2]|

Step 6: Simplify further.
|(an+1) / an| = |(x+2) / (2n+2)|

Step 7: Take the limit of the ratio as n approaches infinity.
lim [(x+2) / (2n+2)] as n approaches infinity = 0

Since the limit of the ratio is less than 1, the series converges for all values of x.

The next step is to find the radius of convergence. The radius of convergence (R) is determined by the following formula:

R = 1 / lim sup |(an+1) / an|

Step 8: Calculate the limit superior.
lim sup |(an+1) / an| = lim sup |(x+2) / (2n+2)| as n approaches infinity

Step 9: Simplify further.
lim sup |(x+2) / (2n+2)| = |x+2| / 2

Therefore, the radius of convergence is R = 1 / (|x+2| / 2) = 2 / |x+2|

Finally, the interval of convergence is determined by the range of x values that satisfy |x+2| < 2. Solving for x, we have:

-2 < x+2 < 2

Subtracting 2 from each part:

-4 < x < 0

Hence, the interval of convergence is -4 < x < 0.