(a) If { an } is convergent, show that limn→∞ an+1 = limn→∞ an (b) A sequence { an } is defined

step 1 So for part A, we're supposing that the sequence is conversion. So let's just go ahead and denot

(a) If { an } is convergent, show that limn→∞ an+1 = limn→∞...

(a) If $ \left \{ a_n \right\} $ is convergent, show that $ \displaystyle\lim_{n\to\infty} a_{n+1} = \displaystyle\lim_{n\to\infty} a_n $ (b) A sequence $ \left\{ a_n \right\} $ is defined by $ a_1 = 1 $ and $ a_{n + 1} = 1/(1 + a_n) $ for $ n \ge 1. $ Assuming that $ \left\{ a_n \right\} $ is convergent, find its limit.

(a) If $ \left \{ a_n \right\} $ is convergent, show that
$ \displaystyle\lim_{n\to\infty} a_{n+1} = \displaystyle\lim_{n\to\infty} a_n $
(b) A sequence $ \left\{ a_n \right\} $ is defined by $ a_1 = 1 $ and $ a_{n + 1} = 1/(1 + a_n) $ for $ n \ge 1. $ Assuming that $ \left\{ a_n \right\} $ is convergent, find its limit.

Transcript

00:02 So for part a, we're supposing that the sequence is conversion.

00:06 So let's just go ahead and denote the limit by some letter s.

00:11 That's the limit of the sequence.

00:15 Now, we'd like to show that if we add one to the end, that this new sequence also converges to the same answer.

00:25 So we'll use the fact that by definition 2 in this section, we can say, for each epsilon bigger than zero, there exist and capital n such that if we take little n to be bigger than big n, then a n minus s is less than epsilon.

01:01 So here we have n plus 1.

01:05 So if n is bigger than n, then n plus 1 is also bigger than n.

01:11 And this implies a n plus 1.

01:16 Minus s is less than epsilon.

01:19 So basically down here, we can go ahead and write that more formally, but that's basically the idea that we'll use here.

01:29 So since n plus 1 is bigger than n, and little n is bigger than capital n by transitivity of inequality, n plus 1 is bigger than capital n.

01:48 So by this condition up here, the distance between a and plus 1 and s must be less than epsilon so this shows that the limit of a n plus 1 is also equal to s and this verifies part a that if you add 1 to the end it does not change the limit now let's go ahead onto the next page for part b so we're given a of 1 equals 1 and you're even given a formula for a n plus 1 assuming that it converges let's find the limit so let's go on to the next page so here and is bigger they're equal to one and we're supposing that the limit of a n exist let's just go ahead and call it s again as we did before so using part a so by part a the limit of a and plus one is also equal to s so if we go ahead and take the limit of both sides one over one 1 plus an.

03:17 So we know this equation.

03:19 This is given.

03:20 All we're doing is taking this equation and applying the limit on both sides.

03:26 On the left -hand side from part a, we know that limit is s.

03:30 And on the right -hand side, this is just 1 over 1 plus s because the limit only applies to a -n, not the ones.

03:41 Now let's take this equation here, multiply out the denominator to both sides.

03:47 Then you get s squared plus s minus 1 equals 0.

03:53 Go ahead and use the quadratic formula here, and you get two answers.

04:06 And i claim that only one of these answers will work.

04:09 You have a positive answer.

04:11 You add a negative answer, and the negative answer is not allowed...