00:00 Intro

00:14 Recalling series

00:42 Geometric series

04:05 Harmonic series

09:05 Credits

1 00:00:00,429 –> 00:00:03,521 Hello and welcome back to real analysis

2 00:00:03,721 –> 00:00:08,866 and as always I want to thank all the nice people that support this channel on Steady or Paypal.

3 00:00:09,229 –> 00:00:12,987 In today’s part 16 we will still talk about series

4 00:00:13,800 –> 00:00:17,886 and we have already learned, a series is just a special sequence.

5 00:00:18,244 –> 00:00:23,151 Namely it’s simply the sequence of partial sums for a given sequence a_k.

6 00:00:23,351 –> 00:00:26,535 Hence the sum ends here with a new index n.

7 00:00:27,043 –> 00:00:34,833 Now because you already know this definition, you see we don’t have a problem when we start a sum with another number, which is not 1.

8 00:00:35,033 –> 00:00:42,953 So depending on the sequence a_k, we can start here with 0, 2 or 3 or even a negative number if it makes sense.

9 00:00:43,153 –> 00:00:48,272 I tell you that because in the next example we have a series that starts with 0.

10 00:00:48,929 –> 00:00:51,843 It’s the very important geometric series

11 00:00:52,043 –> 00:00:56,473 and there we sum all the powers of a given number q.

12 00:00:56,673 –> 00:01:01,019 Hence for us first q could be any real number.

13 00:01:01,219 –> 00:01:09,912 However what we want to show now is that this series is convergent if and only if q is less than 1 in the absolute value.

14 00:01:10,112 –> 00:01:15,926 Hence only in this case we get a finite number, when calculating this infinite sum.

15 00:01:16,414 –> 00:01:19,124 Ok, then let’s start the calculation.

16 00:01:19,324 –> 00:01:25,330 The overall question is of course, what is our sequence s_n? The sequence of partial sums.

17 00:01:25,843 –> 00:01:30,301 Because when we know this, we can calculate the limit n to infinity.

18 00:01:30,501 –> 00:01:36,351 Indeed there is a nice calculation we can do with this sum, in the case that q is not equal to 1.

19 00:01:36,929 –> 00:01:42,867 Of course this is not a restriction, because if q is equal to 1 we have a very simple sum here.

20 00:01:43,067 –> 00:01:47,999 We just sum up 1’s and therefore the series is divergent to infinity.

21 00:01:48,429 –> 00:01:54,893 Ok, now in the case that q is not 1, we can multiply the whole sum with (1 - q).

22 00:01:55,093 –> 00:01:58,734 So we just scale the sequence here with a non-zero number.

23 00:01:58,934 –> 00:02:04,430 Now, the overall idea is that we can expand this expression here and get 2 sums.

24 00:02:04,630 –> 00:02:10,001 The first sum is the same as before and the second has q to the power (k + 1)

25 00:02:10,571 –> 00:02:16,969 and now what we can do is change the index here, such that we have again, q to the power k inside the sum.

26 00:02:17,169 –> 00:02:23,129 This simply means that we start with k=1 and go to n+1

27 00:02:23,329 –> 00:02:27,185 and then you should see, we go exactly through the same numbers.

28 00:02:27,385 –> 00:02:32,976 Ok, such an index shift is very helpful, because now we can subtract the 2 sums here

29 00:02:33,400 –> 00:02:39,364 and the only things that remain are for the first sum, the first index, so k=0

30 00:02:39,564 –> 00:02:44,517 and for the second sum the last index, so k=n+1.

31 00:02:44,929 –> 00:02:51,228 Hence we have q to the power 0, which is by definition 1 and q to the power n+1.

32 00:02:51,428 –> 00:02:55,903 Ok, with this we have a very nice expression for our sequence s_n.

33 00:02:56,103 –> 00:03:02,809 Namely we can just divide by 1-q, which is non-zero and get this nice formula here

34 00:03:03,171 –> 00:03:08,415 and there you should immediately see in which cases s_n is a convergent sequence.

35 00:03:08,615 –> 00:03:13,389 We just have to know if the sequence q to the power n is convergent.

36 00:03:13,589 –> 00:03:17,028 This is what you might already know and it is not hard to show.

37 00:03:17,157 –> 00:03:23,036 q to the power n is convergent if and only if the absolute value of q is less than 1

38 00:03:23,236 –> 00:03:27,464 and in this case we also know the limit, which is 0.

39 00:03:27,664 –> 00:03:37,181 Then this implies that we can actually calculate the limit of this sequence here and as you can see, this is 1 over 1-q.

40 00:03:37,381 –> 00:03:47,542 Hence in summary the value of our series here is well defined for q in the absolute value less than 1 and it is 1/(1-q).

41 00:03:47,742 –> 00:03:51,580 This is such an important formula, such that it gets its own name.

42 00:03:52,143 –> 00:03:56,151 It’s called the geometric series and you really should remember it.

43 00:03:56,351 –> 00:04:01,098 Of course you can see that this formula is important, because it got a name.

44 00:04:01,298 –> 00:04:05,885 Therefore I can now tell you there is another series that has a special name.

45 00:04:06,085 –> 00:04:09,044 This one is the so called Harmonic series

46 00:04:09,300 –> 00:04:13,088 and indeed it’s an example of a divergent series.

47 00:04:13,288 –> 00:04:17,897 It’s constructed by summing up all fractions 1/k.

48 00:04:18,343 –> 00:04:22,630 So we have 1 + 1/2 + 1/3 + 1/4 and so on

49 00:04:23,071 –> 00:04:28,709 and now we will show that this is not convergent, but rather divergent to +infinity.

50 00:04:28,909 –> 00:04:35,581 The first time you see this it might be a little bit surprising, because each number we add gets smaller and smaller.

51 00:04:35,957 –> 00:04:43,143 However the whole sum increases as much as you want and this is exactly what we now want to show.

52 00:04:43,466 –> 00:04:47,382 Therefore let’s look again at the partial sums we call s_n.

53 00:04:47,582 –> 00:04:54,687 Here you should see immediately that this is a sequence of positive real numbers, that is also monotonically increasing.

54 00:04:55,157 –> 00:05:00,097 Hence the only thing that remains to show is that this sequence is not bounded.

55 00:05:00,297 –> 00:05:05,554 In order to show this we have different possibilities, but I choose a very simple one.

56 00:05:05,754 –> 00:05:09,945 We will only look at the Indices, where n is a power of 2.

57 00:05:10,145 –> 00:05:13,037 There you will see, this makes everything easier.

58 00:05:13,237 –> 00:05:17,436 In the first step you should see we can rewrite this number as differences.

59 00:05:17,636 –> 00:05:23,928 This means that I can start with s_1 and then I go to s_2, but then I subtract s_1 again.

60 00:05:24,128 –> 00:05:27,126 Hence this whole term is just s_2.

61 00:05:27,326 –> 00:05:30,232 Then we go to s_4 and subtract s_2 again.

62 00:05:30,929 –> 00:05:37,427 In other words this is just s_4 and please recall I just want to use the powers of 2 here.

63 00:05:37,886 –> 00:05:41,831 Therefore in the next step I would use s_8 and subtract s_4 again

64 00:05:42,714 –> 00:05:47,379 and then we continue the whole game until we reach our index 2 to the power m

65 00:05:47,579 –> 00:05:51,855 and of course in this last step we have to subtract the predecessor.

66 00:05:52,055 –> 00:05:57,600 In summary this is just a complicated way to write s with the index 2 to the power m.

67 00:05:57,800 –> 00:06:02,894 However, still this is very helpful, because now we have differences here

68 00:06:03,094 –> 00:06:06,880 and these differences are much easier to estimate.

69 00:06:07,080 –> 00:06:13,029 Of course if you don’t like the dots here, we can simply put the whole formula into a sum symbol here.

70 00:06:13,229 –> 00:06:19,119 It has the same information, but now it is compact, but we have to use another index j.

71 00:06:19,319 –> 00:06:24,439 Ok, at this point you should recall that s with an index stands for a partial sum.

72 00:06:25,000 –> 00:06:31,996 So here we subtract 2 partial sums and of course what remains is the middle part of the partial sum.

73 00:06:32,500 –> 00:06:35,249 Ok so let’s write that into another line.

74 00:06:35,600 –> 00:06:43,379 Ok, so we know we get out a partial sum, the only thing we have to think about is “what is the start index?” and “what is the end index?”

75 00:06:43,886 –> 00:06:53,202 For this recall that this partial sum wants to the end index 2 to the power j and this end index is smaller than this.

76 00:06:53,671 –> 00:06:57,056 Hence the final index has to be 2 to the power j.

77 00:06:57,256 –> 00:07:03,258 Also we know all indices that are smaller than this or equal will vanish.

78 00:07:03,458 –> 00:07:08,502 Hence the start index has to be 2 to the power (j - 1 + 1).

79 00:07:08,702 –> 00:07:13,313 Of course this looks complicated, because of all the indices involved,

80 00:07:13,513 –> 00:07:19,009 but actually it’s not so complicated, because it’s just a small part of this sum here.

81 00:07:19,209 –> 00:07:25,112 Therefore maybe for the visualization just imagine that the sum stands for these three numbers here.

82 00:07:25,312 –> 00:07:34,972 Then of course the sum gets smaller when we don’t add 1/2 + 1/3 + 1/4, but rather 3 times 1/4 here.

83 00:07:35,172 –> 00:07:42,049 Indeed that’s what we can do in general when we put in here for the fraction the largest index.

84 00:07:42,249 –> 00:07:46,915 Now we just have to count how many times we add this constant number here

85 00:07:47,115 –> 00:07:52,724 and when you do this you see we have exactly 2 to the power j-1 terms.

86 00:07:53,314 –> 00:07:56,439 Hence this is the only product we have to calculate now

87 00:07:56,639 –> 00:08:02,949 and because we have chosen the powers of 2 this is not so complicated. It’s exactly 1/2.

88 00:08:03,400 –> 00:08:11,959 Now this is an important information, because it tells us that this term here is always greater than 1/2 no matter what j is.

89 00:08:12,159 –> 00:08:18,180 So we have the factor 1/2 m-times. In conclusion here we have our estimate.

90 00:08:18,380 –> 00:08:24,959 The partial sum as with index 2 to the power m is always greater than s_1 + m times 1/2

91 00:08:25,514 –> 00:08:31,695 and of course this is not bounded at all, because when m increases this goes to infinity.

92 00:08:32,214 –> 00:08:36,914 Hence we have shown that the whole sequence of partial sums is not bounded from above.

93 00:08:37,457 –> 00:08:42,028 In combination with the monotonicity we have our divergence to infinity.

94 00:08:42,329 –> 00:08:50,410 Hence there we have our harmonic series, which is a divergent series. So please remember also this important result.

95 00:08:50,610 –> 00:08:57,668 Ok then in the next videos we will talk more about criteria we can use to show the convergence of series

96 00:08:57,868 –> 00:09:03,156 and you will also see that the 2 examples we had here are also very helpful there.

97 00:09:03,356 –> 00:09:05,814 So I hope I see you next time. Bye.

Q1: For which $q \in \mathbb{R}$ is the series $\displaystyle \sum_{k = 0}^\infty q^k$ convergent?

A1: Only for $q = 0$

A2: For none.

A3: Only for $|q| < 1$.

A4: Only for $|q| \leq 1$.

A5: Only for $|q| \geq 1$.

Q2: Which statement is correct?

A1: For all $q \in \mathbb{R}$, we have $\displaystyle \sum_{k = 0}^\infty q^k = \frac{1}{1 - q}$

A2: For all $|q| < 1$, we have $\displaystyle \sum_{k = 0}^\infty q^k = \frac{1}{1 - q}$

A3: For all $|q| < 1$, we have $\displaystyle \sum_{k = 0}^\infty q^k = \frac{1}{1 + q}$

A4: For all $|q| > 1$, we have $\displaystyle \sum_{k = 0}^\infty q^k = \frac{2}{1 + q}$

Q3: The harmonic series $\displaystyle \sum_{k = 1}^\infty \frac{1}{k}$ satifies a lot of properties. Which one is not correct?

A1: The harmonic series is divergent.

A2: The harmonic series does not converge.

A3: The harmonic series is strictly monotonically increasing.

A4: The harmonic series is a Cauchy sequence.

A5: The harmonic series is divergent to $\infty$.

A6: The harmonic series is not bounded from above.

A7: The harmonic series is bounded from below.