Question

test the series for convergence or divergence. − 2/ 4 + 4/ 5 − 6/ 6 + 8/ 7 − 10 /8 +....identify bn. (assume the series starts at n = 1.)

Answer 1

To test the convergence or divergence of the given series, we first find the general term, bn, which is (-1)^n * (2n + 2) / (n + 3). Then we split the series into two separate series and evaluate their limits as n approaches infinity. The limits are not the same, indicating that the given series diverges.

Step-by-step explanation:

To test the convergence or divergence of the series, we need to find the limit of the terms of the series as n approaches infinity.

First, let's find the general term, bn. The numerator of each term alternates between negative and positive even numbers, starting with -2 and increasing by 2 in each term. The denominator of each term is n + 3.

So, bn = (-1)^n * (2n + 2) / (n + 3).

Now, let's evaluate the limit of bn as n approaches infinity. Since the numerator has alternating signs, we can split the series into the sum of two separate series: Σ((-1)^n * 2n) / (n + 3) and Σ((-1)^n * 2) / (n + 3).

The limit of Σ((-1)^n * 2n) / (n + 3) as n approaches infinity is 2.

The limit of Σ((-1)^n * 2) / (n + 3) as n approaches infinity is 0.

Since the limit of the two series is not the same, the given series diverges.

Answer 2

The given series is an alternating series and converges according to the Alternating Series Test, as the terms decrease monotonically and approach zero.

Step-by-step explanation:

To test the series for convergence or divergence, the given series is − 2/4 + 4/5 − 6/6 + 8/7 − 10/8 + .... This appears to be an alternating series where the nth term can be represented as −(2n)/(n+3) when n is odd, and (2n)/(n+3) when n is even. A general term for the series can thus be defined using bn = −(2n)1 - 2(n mod 2)/(n+3), where 'n mod 2' provides 0 for even n and 1 for odd n.

To determine convergence, we can apply the Alternating Series Test. We observe that the absolute value of the terms |bn| = 2n/(n+3), which simplifies to − 2/n for large n, is monotonically decreasing and approaches zero as n approaches infinity. Since both conditions of the Alternating Series Test are met, the series converges.