
Title: Example Calculation

Series: Real Analysis

Chapter: Sequences and Limits

YouTubeTitle: Real Analysis 8  Example Calculation

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Quiz: Test your knowledge

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Subtitle on GitHub: ra08_sub_eng.srt

Timestamps
00:00 Intro
00:21 Recalling the monotone convergence criterion
00:55 Introducing the example
01:21 Proving monotonicity
04:49 Proving bounded from above
08:55 End result of the example
09:46 Credits

Subtitle in English
1 00:00:00,371 –> 00:00:03,510 Hello and welcome back to real analysis
2 00:00:04,214 –> 00:00:09,751 and as always many, many thanks to all the nice people that support this channel on Steady or Paypal.
3 00:00:10,471 –> 00:00:15,429 This part 8 today is about an example, where we can do some concrete calculations.
4 00:00:15,800 –> 00:00:21,152 In particular i want to show you how we can apply this fact, we introduced in the last video.
5 00:00:21,871 –> 00:00:26,153 This is one important criterion you can use to check for convergence.
6 00:00:26,571 –> 00:00:32,272 You just have to show that the sequence is monotonically decreasing and also bounded from below.
7 00:00:32,686 –> 00:00:35,385 In a similar way this works the other way around.
8 00:00:35,585 –> 00:00:43,442 So if you can show that the sequence is monotonically increasing and also bounded from above, then the sequence is also convergent.
9 00:00:44,314 –> 00:00:48,646 Now, both things we can just call the “monotone convergence criterion”.
10 00:00:49,371 –> 00:00:54,415 Ok, now the video is about how to use this criterion for an example.
11 00:00:55,057 –> 00:01:00,697 Indeed this example is a very important one, because it has a very special limit.
12 00:01:01,014 –> 00:01:08,065 So here the sequence a_n is given by “(1 + 1/n) to the power n”.
13 00:01:08,786 –> 00:01:14,445 In order to show that this is indeed a convergent sequence, we can use the criterion from above.
14 00:01:15,114 –> 00:01:20,234 Hence we have to show 2 seperate properties and we start with the monotonicity.
15 00:01:20,757 –> 00:01:28,613 Ok, there i can give you the hint: often it is helpful to consider the fraction “a_n+1 divided by a_n”
16 00:01:28,813 –> 00:01:35,131 and then you can see if this is less or equal than 1, this sequence is monotonically decreasing.
17 00:01:35,914 –> 00:01:41,873 On the other hand if the fraction is greater or equal than 1, this sequence is monotonically increasing.
18 00:01:42,414 –> 00:01:45,770 So for this reason lets analyse this fraction.
19 00:01:46,071 –> 00:01:55,137 In the first step by just putting in the definition, we immediately see we have in the numerator the power n+1 and in the denominator the power n.
20 00:01:55,337 –> 00:01:58,602 Therefore the first idea would be to adjust that.
21 00:01:59,129 –> 00:02:06,028 One possibility would be to increase the power in the denominator, but then we have to cancel this out with the same factor.
22 00:02:06,829 –> 00:02:10,094 Ok, with this you should see we didn’t change anything.
23 00:02:10,857 –> 00:02:14,781 Then in the next step i want to pull the power out of the fraction,
24 00:02:15,557 –> 00:02:23,077 because now we can easily work inside the parentheses and get rid of all the fractions inside the large fraction.
25 00:02:23,657 –> 00:02:28,641 In order to do this we just expand the fraction by a factor “n times (n+1)”.
26 00:02:29,471 –> 00:02:33,020 There you can just do the multiplication and then we get the following.
27 00:02:33,220 –> 00:02:37,129 In the numerator we have “n times (n +1) + n”
28 00:02:37,244 –> 00:02:41,171 but in the denominator we also have + 1 at the end.
29 00:02:41,857 –> 00:02:47,513 Therefore we can just add the missing 1 in the numerator as well, if we subtract it immediately.
30 00:02:47,929 –> 00:02:53,322 Of course the advantage now is that we can split this fraction into 2 simple fractions.
31 00:02:53,900 –> 00:02:57,844 The first is just the number 1, because it nicely cancels out
32 00:02:58,044 –> 00:03:01,499 and the second is 1 divided by the denominator.
33 00:03:01,943 –> 00:03:07,925 Indeed simplifying this denominator gets us “n^2 + 2n + 1”
34 00:03:08,125 –> 00:03:11,898 and you should know this is simply (n + 1) squared.
35 00:03:12,714 –> 00:03:17,062 Now in the next step we use something that is called “Bernoulli’s inequality”.
36 00:03:17,571 –> 00:03:24,829 It tells us that for natural numbers k and real numbers greater or equal than 1 we have
37 00:03:25,200 –> 00:03:31,520 that we can transform the power k to a multiplication and then we get something that is less or equal.
38 00:03:31,957 –> 00:03:35,802 This statement here you can easily show by using induction
39 00:03:36,514 –> 00:03:40,702 and then we can use it to estimate any powers we have of this form.
40 00:03:41,343 –> 00:03:47,498 In our case we use it for this factor here, where the power is n + 1 and this is our x.
41 00:03:47,698 –> 00:03:55,777 Hence the whole thing is greater or equal than when we have here, “1 + (n + 1) times our x”.
42 00:03:56,286 –> 00:03:59,842 Which is 1 divided by (n + 1) squared.
43 00:04:00,571 –> 00:04:03,741 Hence you see this (n+1) here cancels out.
44 00:04:04,529 –> 00:04:08,308 Ok, after this we can just put both factors here together.
45 00:04:08,657 –> 00:04:13,974 The first factor we can just rewrite as (n + 1) divided by n.
46 00:04:14,600 –> 00:04:20,235 and in the same way the second factor is “(n + 1)  1” divided by (n + 1).
47 00:04:20,900 –> 00:04:26,408 and there you see in the numerator we have + 1 and 1, so in sum just 0.
48 00:04:27,286 –> 00:04:30,724 Hence we can just cancel all the factors here.
49 00:04:31,486 –> 00:04:34,695 Which leads us to a simple number 1.
50 00:04:35,400 –> 00:04:41,776 Therefore with this we have shown that this fraction here is greater or equal than 1.
51 00:04:42,386 –> 00:04:48,837 This holds for all n. So we have shown that the sequence a_n is indeed monotonically increasing.
52 00:04:49,443 –> 00:04:54,437 Hence in the second part now, we show that the second is also bounded from above.
53 00:04:55,014 –> 00:05:00,563 Showing this is indeed a good exercise, because i can explain the binomial theorem.
54 00:05:01,086 –> 00:05:04,567 It tells us that we can write this binomial as a sum.
55 00:05:05,214 –> 00:05:09,603 The sum starts for k = 0 and goes until n.
56 00:05:09,803 –> 00:05:13,782 Then comes the binomial coefficient n over k
57 00:05:14,371 –> 00:05:20,814 and then the first entry to the power (n  k) times the second entry to the power k.
58 00:05:21,171 –> 00:05:26,713 Hence in our case this looks very simple, because 1 stays 1, no matter what the power is.
59 00:05:27,329 –> 00:05:32,133 Ok in order to get an idea what this sum means, lets calculate the first 2 terms here.
60 00:05:32,643 –> 00:05:40,701 So we have n choose 0 as a binomial coefficient times 1 to the power n times 1/n to the power 0.
61 00:05:41,129 –> 00:05:45,440 Then the second term is the same with k = 1
62 00:05:45,943 –> 00:05:49,315 and then for all the other terms we just write a sum again.
63 00:05:49,629 –> 00:05:53,139 Ok, then lets discuss all the different factors we have here.
64 00:05:53,886 –> 00:05:58,487 Now by definition you should know that n choose 0 is just 1
65 00:05:58,687 –> 00:06:01,424 and the same by definition for the power 0.
66 00:06:02,171 –> 00:06:05,342 Therefore the first part in the sum is just the number 1.
67 00:06:05,986 –> 00:06:10,700 In the second part we have n choose 1, which is by definition n
68 00:06:11,114 –> 00:06:14,525 and also 1/n. So they cancel out.
69 00:06:14,900 –> 00:06:19,365 In other words also for the second part in the sum we have the number 1.
70 00:06:20,043 –> 00:06:24,756 Hence we only need to discuss what happens with the rest of the sum. So the third part here.
71 00:06:25,414 –> 00:06:30,105 For this it might be helpful to recall the definition of the binomial coefficient.
72 00:06:30,514 –> 00:06:39,512 Indeed n choose k is given by n! divided by “(n  k)! times k!.
73 00:06:40,114 –> 00:06:46,191 Now the idea is that we combine this with the second factor, which is 1/n to the power k.
74 00:06:46,729 –> 00:06:49,832 Therefore lets put these three factors together.
75 00:06:50,243 –> 00:06:56,880 Now n! is nothing else then “n times (n  1) times (n  2) and so on.
76 00:06:57,080 –> 00:07:05,058 However, if we divide it by (n  k)!, we reach an end point here. Which is (n  k + 1).
77 00:07:05,700 –> 00:07:13,981 Now what you should see is, that we have exactly k factors here. Therefore n to the power k, we also can write as a product.
78 00:07:14,471 –> 00:07:18,562 Then each factor in the numerator gets an n in the denominator
79 00:07:18,914 –> 00:07:24,718 and then we can conclude that each pair we put together here is less or equal than 1.
80 00:07:25,557 –> 00:07:28,929 Therefore the whole fraction here is less or equal than 1.
81 00:07:29,743 –> 00:07:36,541 Hence we have a nice estimate. We can just say that the whole thing is less or equal than 1/k!
82 00:07:37,043 –> 00:07:44,073 and because we started a sum for k = 2, we can also estimate this 1/k!.
83 00:07:44,614 –> 00:07:48,346 We just stop the factorial product after 2 steps,
84 00:07:48,546 –> 00:07:53,298 because we divide by a smaller number than, we just get out a bigger number.
85 00:07:54,057 –> 00:07:58,257 Then in the last step we just expand this fraction into 2 fractions.
86 00:07:59,086 –> 00:08:02,810 We simply have 1/(k  1)  1/k.
87 00:08:03,400 –> 00:08:06,211 Please check that this is indeed equal.
88 00:08:06,411 –> 00:08:11,227 Ok, then calculating the sum for these 2 fractions is very simple.
89 00:08:11,427 –> 00:08:15,061 Because this is what some people call a telescoping sum.
90 00:08:15,714 –> 00:08:20,995 You see there is just one index shift and therefore in the sum a lot of things cancel out.
91 00:08:21,529 –> 00:08:29,264 Indeed only 2 things remain, namely the first part here for k = 2 and the last part here for k = n.
92 00:08:29,464 –> 00:08:33,055 In other words we have 1  1/n.
93 00:08:33,255 –> 00:08:38,937 Ok, now putting all these things together, we can use it to estimate our a_n here.
94 00:08:39,814 –> 00:08:44,613 So it’s less or equal than 2 + 1  1/n.
95 00:08:45,357 –> 00:08:49,622 In summary a_n is less or equal than 3.
96 00:08:50,100 –> 00:08:54,283 Therefore in fact the sequence a_n is bounded from above.
97 00:08:55,000 –> 00:09:00,758 Ok and together with the monotonicity we have that the sequence is actually convergent
98 00:09:01,357 –> 00:09:05,250 and there you see how helpful the monotone convergence criterion can be.
99 00:09:06,271 –> 00:09:12,854 We didn’t need to calculate the limit, we just showed that the sequence is convergent without using the limit.
100 00:09:13,386 –> 00:09:18,135 However now we know that this limit exists and it gets a special name.
101 00:09:18,335 –> 00:09:22,288 It’s simply called e, which stands for Euler’s number
102 00:09:23,014 –> 00:09:28,505 and indeed the whole proof showed here, that Euler’s number has to be between 2 and 3,
103 00:09:29,214 –> 00:09:33,509 but of course about the number e we will talk a lot in this course later.
104 00:09:34,114 –> 00:09:38,937 However before we do this we need to explain more details about sequences first
105 00:09:39,529 –> 00:09:42,253 and that’s what we do in the next video.
106 00:09:42,453 –> 00:09:46,015 Therefore i hope i see you there and have a nice day. Bye!

Quiz Content
Q1: A sequence $(a_n){n\in \mathbb{N}}$ with $$\frac{a{n+1}}{a_n} \geq 1 \text{ for all } n \in \mathbb{N}$$ is always
A1: strictly monotonically increasing.
A2: monotonically increasing.
A3: strictly monotonically decreasing.
A4: monotonically decreasing.
Q2: A sequence $(a_n){n\in \mathbb{N}}$ with $$ a{n+1}  a_n \geq 0 \text{ for all } n \in \mathbb{N}$$ is always
A1: strictly monotonically increasing.
A2: monotonically increasing.
A3: strictly monotonically decreasing.
A4: monotonically decreasing.
Q3: The sequence $ \left( (1 + \frac{1}{n})^n \right)_{n \in \mathbb{N}} $ is monotonically increasing and bounded from above by $3$. What is the correct conclusion?
A1: The sequence is divergent.
A2: The sequence is convergent with limit $3$.
A3: The sequence is convergent with limit greater than $3$.
A4: The sequence is convergent with limit less or equal than $3$.
A5: The sequence is not bounded from below.
Q4: What is the correct Bernoulli’s inequality?
A1: For all $k \in \mathbb{N}$ and $x \geq 1$, we have $$ (1+x)^k \leq 1 + k x , .$$
A2: For all $k \in \mathbb{N}$ and $x \geq 1$, we have $$ (1+x)^k < 1 + k x , .$$
A3: For all $k \in \mathbb{N}$ and $x \geq 1$, we have $$ (1+x)^k \geq 1 + k x , .$$
A4: For all $k \in \mathbb{N}$ and $x \geq 1$, we have $$ (1+x)^k > 1 + k x , .$$