for each series determine if the series is absolutely convergent and convergent

the sum from 0 to infinity of (-1)^n/(the square root of (n+1))

I did the ratio test and got -1, which is less than 0 making it absolutely convergent, but do i need to take the absolute value of -1? which would then make the test inconclusive?

Yes, you need to take the absolute value of everything, so you get 1 which is inconclusive. If you do the integral test, you find that it diverges.

To determine whether the series is absolutely convergent or not, we need to check if the series formed by taking the absolute value of each term converges.

In this case, the series is given by:

S = Σ (-1)^n / √(n+1), where n starts from 0 to infinity.

To apply the ratio test, we consider the absolute value of the terms:

|(-1)^n / √(n+1)| = 1/√(n+1).

Now, let's use the ratio test:

lim (n→∞) |(1/√(n+2)) / (1/√(n+1))|.

Simplifying the expression, we get:

lim (n→∞) √(n+1) / √(n+2).

To evaluate this limit, we multiply the numerator and denominator by the conjugate of the numerator:

lim (n→∞) ( √(n+1) / √(n+2) ) * ( √(n+1) / √(n+1) ) = lim (n→∞) (n+1) / (n+2).

As n tends to infinity, the limit evaluates to 1. Since the limit is less than 1, the series is absolutely convergent.

Therefore, you do not need to take the absolute value of -1 in this case. The ratio test confirms that the series is absolutely convergent.