Select the FIRST correct reason why the given series...

Select the FIRST correct reason why the given series converges. A. Convergent geometric series B. Convergent $p$ series C. Comparison (or Limit Comparison) with a geometric or $p$ series D. Alternating Series Test E. None of the above 1. $\sum_{n=1}^{\infty} \frac{cos(n\pi)}{ln(6n)}$ 2. $\sum_{n=1}^{\infty} \frac{(-1)^n ln(e^n)}{n^8 cos(n\pi)}$ 3. $\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n+7}$ 4. $\sum_{n=1}^{\infty} \frac{(n+1)(80)^n}{9^{2n}}$ 5. $\sum_{n=1}^{\infty} \frac{3(8)^n}{10^{2n}}$ 6. $\sum_{n=1}^{\infty} \frac{sin^2(3n)}{n^2}$

Transcript

00:01 Ok, we're going to go through each of these, and i'm going to go ahead and tell you the first of the reasons a, b, c, d, or e, given that the series does converge.

00:10 So 1, 2, 3, 4, 5, and 6.

00:12 The first series, we have a negative 1 to the end that indicates an alternating series.

00:17 Given that, the answer for that one is d.

00:20 It will converge by the alternating series test because it is decreasing, and its limit does go to 0.

00:26 Our next question here is part 2.

00:29 That one's going to be a geometric series, because the ratio there is going to be 5 over 49, which does converge.

00:36 So the answer to that one is a.

00:39 Number 3, sine squared of 2n over n squared.

00:42 This is one where you're going to do an actual comparison, because you know that sine squared has to be between 0 and 1.

00:49 So you can note that sine squared of 2n over n squared has to be less than or equal to 1 over n squared.

00:56 And then this is a comparison.

00:58 Right? and by a comparison, this does converge because of p series.

01:03 So since the larger function converges, the smaller one also does.

01:07 So this one would be c.

01:10 Number 4, negative 1 to the n ln e to the n, and then divided by all that junk.

01:15 That is going to converge, because it's an alternating series.

01:18 We have negative 1 to the n.

01:19 That's kind of a dead giveaway...

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