1. (8 points) Write the first four terms of each...

1. (8 points) Write the first four terms of each sequence. (a) {u221an / (n + 1)}_{n=0}^u221e (b) a_n = (2n - 1) / n^2 (c) a_n = sin(u03c0n / 2) (d) {(2 + (-1)^n) / (3n - 1)} 2. (8 points) Find the following limits. Show all work and use appropriate notation. (a) lim_{nu2192u221e} (ln n / u221an) (b) lim_{nu2192u221e} (5^n / n^n)

Transcript

00:01 Okay, so for this problem, we are writing out the first four terms of some of the sequences.

00:07 So the first sequence is square root of n over n plus 1.

00:16 So this is another way of writing a sequence, denoting the general term of a sequence.

00:23 So n starts at 0 and n goes infinity.

00:26 But we're writing the first four terms, so n starts at 0, you're going to have square root of 0 over 1.

00:32 And then the next one will be square root of 1 over 2 and then square root of 2 over 3 and square root of 3 over 4 so simplifying this and this is just the first 4 term so we get 0 1 1⁄2 over 3 and square root of 3 over 4 for the second sequence you have this notation a of n equals 2 n n minus 1 over n square.

01:13 So for this one, it doesn't tell you where, which number n starts at, which integer n starts at, whether it's 0, 1, but you should know you can't have n equal 0 because you can divide by 0.

01:27 So we'll make n starts at 1.

01:29 So 2 times 1 is 2, 2 minus 1 is 1 over 1 square.

01:33 And then the next term would be 2.

01:36 I mean the next end would be 2, so 4 minus 1.

01:39 Minus 1 is 3 over 2 square, that's 4.

01:43 And then 2 times 3 is 6 minus 1 is 5 over 3 square is 9.

01:51 So notice how, because of the 2, these numerators go up by 2.

01:57 So next one should be 7.

01:58 2 times 4 minus 1, and then you have square of 4, which is 16.

02:04 So 1, 3, 4, 5 over 9, 7 over 16.

02:11 That is the second sequence, the first four terms.

02:16 Okay, next part, i mean next sequence, we have a fn equals sine pi n over 2.

02:34 So this one, we can start at zero.

02:38 We can start at 1, but i always start at 0.

02:41 If you can start at 0, if you can start, then, you know, you can start at 0.

02:47 So, sine of 0 pi over 2 is sine of 0, which is 0.

02:53 And then you have sine of 1 pi over 2, which is sine of pi over 2, that's 1.

02:58 And then sign of 2 pi over 2, that's pi.

03:01 So sign of pi is 0 again.

03:04 Just think of the unit circle.

03:06 And then sign of 0 1, 2, 3 pi over 2 is negative 1.

03:14 So those are your first 4 terms.

03:19 Okay, and then the last sequence is a of n equal, oh, it doesn't have a of n in it, just has this bracket, 2 plus negative 1 to the n over 3n minus 1.

03:42 So you can't slide 0 for this, definitely, so because when you plug in 0, you don't have 0 in the denominator, so it's fine.

03:50 So you have 2 plus negative 1 to the 0 over 3 times 0 minus 1.

03:58 There's a lot going on here, so i'm just writing it out.

04:03 2 plus negative 1 to the 1 over 3 times 1 minus 1.

04:12 2 plus negative 1 to the 2 2 2 2 2 2 times 1 to the 2 2 times 1.

04:18 Oh, wait.

04:20 We have to go up an n.

04:23 So it's 3 times 2 minus 1.

04:28 And finally, the fourth term is 2 plus negative 1 to the third power over 3 times 3 minus 1.

04:38 So we have 2 plus 1, negative 1 to 0 is 1...

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