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These are the user uploaded subtitles that are being translated: 1 00:00:00,460 --> 00:00:04,710 In this video we're talking about how to use the alternating series tests to say whether or not a series 2 00:00:04,710 --> 00:00:05,540 converges. 3 00:00:05,670 --> 00:00:09,990 And before we go any further it's really important to remember that the alternating series test can 4 00:00:09,990 --> 00:00:11,570 never prove divergence. 5 00:00:11,580 --> 00:00:15,330 So if you're going to use the alternating series Test the only conclusions you're going to be able to 6 00:00:15,330 --> 00:00:20,640 draw are that the series converges or that the test is inconclusive and you're going to have to try 7 00:00:20,640 --> 00:00:24,880 to use a different test to determine convergence or divergence. 8 00:00:24,900 --> 00:00:28,560 So with the alternating series Test it's convergence or nothing. 9 00:00:28,740 --> 00:00:33,930 So in this particular problem we've been given the series given by some from equals one to infinity 10 00:00:34,320 --> 00:00:41,430 of negative one to the end plus one multiplied by squared of and plus one minus square root of n. 11 00:00:41,430 --> 00:00:47,790 So this negative 1 to the N plus 1 term tells us that this is an alternating series of negative 1 to 12 00:00:47,790 --> 00:00:51,680 the end plus 1 and minus 1 or even just to the end. 13 00:00:51,810 --> 00:00:55,650 This term is going to make the entire series alternate sign. 14 00:00:55,650 --> 00:00:59,090 So it's got alternate positive negative positive negative positive negative. 15 00:00:59,220 --> 00:01:01,050 So it's an alternating series. 16 00:01:01,050 --> 00:01:05,910 The important thing to know about an all training series and for the alternating series tests is that 17 00:01:06,000 --> 00:01:11,890 really most of the time we're just focused on the value outside of this negative one term. 18 00:01:12,150 --> 00:01:16,700 So we're really going to be looking at is this value right here. 19 00:01:16,830 --> 00:01:23,190 And when it comes to the alternating series tests we say for an altering series a seven and a sub and 20 00:01:23,190 --> 00:01:29,100 the series a seven is going to be this value we've put brackets around so everything but this negative 21 00:01:29,100 --> 00:01:31,110 one term that makes the series alternate. 22 00:01:31,170 --> 00:01:38,090 So we drop everything but this negative one term for A7 and what we say is that if the series is decreasing 23 00:01:38,100 --> 00:01:44,820 that's what this inequality means that a sub and plus one is less than a seben and they're both greater 24 00:01:44,820 --> 00:01:45,950 than zero. 25 00:01:46,110 --> 00:01:52,950 So both the series seven plus one and a seven are positive but the Southern plus one term is less than 26 00:01:52,950 --> 00:01:54,000 the seven term. 27 00:01:54,000 --> 00:01:56,100 So the series is decreasing. 28 00:01:56,130 --> 00:02:02,700 So if we can show that and we can show that the infinite limit of a sedan is equal to zero then by the 29 00:02:02,700 --> 00:02:06,110 alternating series Test we can prove that a converges. 30 00:02:06,120 --> 00:02:08,820 So we're really just tackling these two conditions here. 31 00:02:08,910 --> 00:02:14,510 So let's go ahead and start with this inequality that says A7 plus one has to be less than a seven. 32 00:02:14,520 --> 00:02:17,470 There's a couple of ways to do this and I'll show you both of them. 33 00:02:17,490 --> 00:02:23,480 The first one is just to expand the first few terms of the series and compare them across both. 34 00:02:23,520 --> 00:02:30,480 So if we go ahead and say that a seben is going to be equal to the square root of and plus 1 minus the 35 00:02:30,480 --> 00:02:36,650 square root of and we're just taking that directly from our original series here so that's a seven then 36 00:02:36,730 --> 00:02:39,720 a sub and plus 1 so a then plus 1. 37 00:02:39,740 --> 00:02:43,810 What we want to do is replace every value with an end plus 1 value. 38 00:02:43,950 --> 00:02:47,160 So instead of n right here we put an end plus 1. 39 00:02:47,220 --> 00:02:48,870 We keep the other plus 1. 40 00:02:48,870 --> 00:02:54,810 So we've got that there and then we just say minus the square root of and plus 1 because we replace 41 00:02:54,900 --> 00:02:56,730 n with end plus 1. 42 00:02:56,730 --> 00:03:04,220 So what this simplifies to then is the square root of and plus 2 minus the square root of and plus 1. 43 00:03:04,350 --> 00:03:07,250 So this is the series a seven plus one. 44 00:03:07,260 --> 00:03:09,180 This is the series a Sabrine. 45 00:03:09,180 --> 00:03:10,940 So if we start plugging in some values. 46 00:03:10,950 --> 00:03:15,600 Notice that the index of the series starts at 1 so end starts at 1. 47 00:03:15,600 --> 00:03:22,230 So if we start plugging in values for an equals 1 and equals 2 3 4 etc. here's or we're going to get. 48 00:03:22,230 --> 00:03:24,980 So plugging in one we get one plus one is two. 49 00:03:25,020 --> 00:03:31,290 So we get the square root of two minus the square root of 1 because we plug one in here then we're going 50 00:03:31,290 --> 00:03:35,120 to add to that whatever we get when we plug in N equals 2. 51 00:03:35,280 --> 00:03:39,600 So here we'll get the square root of 3 minus the square root of 2. 52 00:03:39,780 --> 00:03:44,580 Then we're going to add to that plugging in here three we're going to get the square root of 4. 53 00:03:44,640 --> 00:03:47,920 Minus the square root of 3 and we could keep going. 54 00:03:48,090 --> 00:03:51,050 Let's look at that same process for a seven plus one. 55 00:03:51,060 --> 00:03:53,320 So here plugging in N equals 1. 56 00:03:53,370 --> 00:03:58,180 We're going to get the square root of 3 minus the square root of 2. 57 00:03:58,530 --> 00:04:00,480 Then plugging in equals two. 58 00:04:00,540 --> 00:04:04,750 Get the square root of four minus the square root of three. 59 00:04:04,920 --> 00:04:11,300 And then pugging in any three will get the square root of five minus the square root of four. 60 00:04:11,370 --> 00:04:17,610 And both of these would continue on indefinitely as we plugged in N equals 4 5 6 7 et cetera. 61 00:04:17,640 --> 00:04:23,760 But what we see happening here is interesting because if we look at this a 7 series we notice that we 62 00:04:23,760 --> 00:04:25,430 have a positive root too. 63 00:04:25,590 --> 00:04:29,310 And then later on in the end equals to term a negative root to. 64 00:04:29,480 --> 00:04:31,270 Here we've got positive root 3. 65 00:04:31,410 --> 00:04:34,010 And then here in the end equals three term a negative root 3. 66 00:04:34,200 --> 00:04:37,520 And that pattern would continue over in the end equals four term. 67 00:04:37,530 --> 00:04:40,140 We would end up with a negative root for term. 68 00:04:40,140 --> 00:04:42,610 So here we're going to get these values to cancel. 69 00:04:42,660 --> 00:04:47,100 We're going to get these values to cancel the square root of four term would cancel with a negative 70 00:04:47,100 --> 00:04:47,910 square root of 4. 71 00:04:47,910 --> 00:04:55,440 Later on down the road the only thing that's going to remain then is this negative square root of 1. 72 00:04:55,470 --> 00:04:57,890 We know that the square root of 1 is one. 73 00:04:57,900 --> 00:05:05,570 So the result of this it's just a negative one here if we look at a seven plus one what we see is a 74 00:05:05,570 --> 00:05:09,560 similar pattern so we're going to get squared of three cancell with negative fruit 3. 75 00:05:09,620 --> 00:05:13,850 We're going to get squared of four to cancel with negative fruit for the square to five would cancel 76 00:05:13,850 --> 00:05:16,330 with a term later on down the line. 77 00:05:16,520 --> 00:05:21,270 So all we're left with then is this negative square root of two value. 78 00:05:21,500 --> 00:05:28,130 So what we could say is that the series is going to be equal to negative route to well negative one 79 00:05:28,130 --> 00:05:31,080 is greater than negative square root 2. 80 00:05:31,100 --> 00:05:35,500 So that shows that a seben is greater than a seven plus one. 81 00:05:35,510 --> 00:05:38,670 You're not always going to get the series to work out that cleanly though. 82 00:05:38,840 --> 00:05:44,690 So the safest way to make sure that the series is decreasing is to actually take the derivative of the 83 00:05:44,690 --> 00:05:46,460 original series. 84 00:05:46,460 --> 00:05:51,950 Remember that the original series is this squared of N plus 1 minus square root of n. 85 00:05:51,980 --> 00:05:58,160 So if we say Y is equal to squared of N plus 1 minus squared even then if we wanted to take the derivative 86 00:05:58,490 --> 00:06:05,780 we would get y prime or d y over d n is going to be equal to and then taking the derivative of this 87 00:06:05,840 --> 00:06:06,630 series. 88 00:06:06,630 --> 00:06:12,140 Remember that the square root of plus 1 is the same thing as quantity and plus 1 race to the one half 89 00:06:12,140 --> 00:06:12,870 power. 90 00:06:13,100 --> 00:06:16,130 So if we want to take the derivative we just use power rule. 91 00:06:16,130 --> 00:06:22,370 We bring that one half ex-porn him down in front and we get 1 1/2 times and plus 1 and then subtracting 92 00:06:22,370 --> 00:06:25,100 1 from the exponent we get a negative 1 1/2. 93 00:06:25,130 --> 00:06:29,520 Same thing year the squirt of end is the same as and to the one half power. 94 00:06:29,540 --> 00:06:33,200 So we bring that one half down in front and we get minus one half. 95 00:06:33,230 --> 00:06:35,220 And to the negative 1 1/2. 96 00:06:35,240 --> 00:06:41,510 Now if we simplify this what we end up with for this first term is 1 divided by two times the square 97 00:06:41,510 --> 00:06:47,990 root of plus 1 because this negative one half becomes a positive one half when we move this and plus 98 00:06:47,990 --> 00:06:49,930 one to the denominator. 99 00:06:50,120 --> 00:06:53,770 So we have and plus one to the positive one half in the denominator. 100 00:06:53,840 --> 00:06:58,880 But of course and plus one to the positive one half is the same as squared of plus 1 which is how we 101 00:06:58,880 --> 00:07:04,610 get that square root in the denominator and then same thing here we would end up with a minus 1 over 102 00:07:04,940 --> 00:07:07,410 two times the square root of N. 103 00:07:07,430 --> 00:07:13,040 Now in order to find a common denominator here we would want to multiply this first fraction by the 104 00:07:13,040 --> 00:07:19,130 squared of an over the square of n we would want to multiply the second fraction by the square root 105 00:07:19,220 --> 00:07:22,870 of plus 1 over the square root of and plus 1. 106 00:07:23,090 --> 00:07:28,550 Because when we do that here in the numerator we'll get square root of N then we'll have this minus 107 00:07:28,640 --> 00:07:34,850 one times square and plus 1 or just minus square root of and plus 1 in the denominator. 108 00:07:34,850 --> 00:07:42,050 Now we have a common denominator of two times the squared of and times the square root of and plus 1. 109 00:07:42,050 --> 00:07:45,720 If we then multiply by the conjugate of the numerator. 110 00:07:45,770 --> 00:07:50,570 Remember that the conjugate of the numerator is the same two terms but with the sign in the middle changed 111 00:07:50,570 --> 00:07:52,880 from a negative to a positive in this case. 112 00:07:52,880 --> 00:07:58,210 So squared of N plus square root of and plus 1 instead of minus. 113 00:07:58,370 --> 00:08:03,230 And then the same thing in the denominator because if the numerator and denominator were different we 114 00:08:03,230 --> 00:08:05,150 wouldn't be multiplying by 1. 115 00:08:05,180 --> 00:08:09,890 So we would be changing the value but here because the numerator and denominator are the same. 116 00:08:09,950 --> 00:08:13,430 This is just like multiplying by 1 so we're not actually changing the value. 117 00:08:13,670 --> 00:08:18,700 But what happens then when we multiply by the conjugate is we get squared event times square to then 118 00:08:18,980 --> 00:08:19,280 which is. 119 00:08:19,280 --> 00:08:23,040 And here we get squared even time squared of N plus 1. 120 00:08:23,150 --> 00:08:27,080 But then we have minus the same thing squared of end time squared event plus 1. 121 00:08:27,080 --> 00:08:33,190 So those middle terms cancel and we're just left with minus the squared of and plus 1 times the squared 122 00:08:33,200 --> 00:08:37,220 of and plus 1 or minus and plus 1. 123 00:08:37,220 --> 00:08:44,840 And then in our denominator if we multiply this two and times route and plus 1 times root n we'll get 124 00:08:44,840 --> 00:08:47,740 the root and to cancel with the root and giving us just. 125 00:08:47,780 --> 00:08:52,070 And so we have two N times the squared of N plus 1. 126 00:08:52,250 --> 00:08:58,830 Then we have a plus to square root n times and plus 1. 127 00:08:58,910 --> 00:09:04,990 If we distribute this negative sign in the numerator what we end up with is and minus and minus 1. 128 00:09:04,990 --> 00:09:09,630 So let's go ahead and change this whole thing to and minus and minus 1. 129 00:09:09,740 --> 00:09:12,860 That will allow us to cancel our ends we have and minus. 130 00:09:12,860 --> 00:09:20,500 And so what we're left with then finally is just a negative one divided by two and square root of end 131 00:09:20,540 --> 00:09:26,800 plus 1 plus two times and plus one Times Square root of n. 132 00:09:26,810 --> 00:09:28,380 So now here's what we see. 133 00:09:28,550 --> 00:09:31,330 We know that the index for an starts at 1. 134 00:09:31,580 --> 00:09:35,770 Which means here that the lowest value the end can possibly have is 1. 135 00:09:35,870 --> 00:09:41,780 It can have values of 1 2 3 4 all the way on up to infinity but it's always going to be a positive number 136 00:09:41,960 --> 00:09:48,470 which means that 2 times n is always going to be a positive number and plus one is always going to be 137 00:09:48,470 --> 00:09:53,300 a positive number and taking the square root of a positive number will always give us a positive number 138 00:09:53,750 --> 00:09:56,350 and plus one will always be a positive number. 139 00:09:56,480 --> 00:10:00,020 And when we multiply that by 2 we'll still always get a positive. 140 00:10:00,310 --> 00:10:03,590 And the square root of N here will always be a positive number. 141 00:10:03,670 --> 00:10:07,850 So we have a positive times a positive plus a positive times a positive. 142 00:10:07,870 --> 00:10:13,070 So it's safe to say that the denominator of this fraction will always be positive. 143 00:10:13,090 --> 00:10:16,190 On the other hand the numerator is a constant negative one. 144 00:10:16,360 --> 00:10:19,820 So the numerator is always going to be a negative number. 145 00:10:20,110 --> 00:10:25,890 So what we can say then is that the derivative D-y dn the derivative is going to be a negative divided 146 00:10:25,890 --> 00:10:26,820 by a positive. 147 00:10:26,900 --> 00:10:29,620 The result of which will always be a negative. 148 00:10:29,740 --> 00:10:32,020 So the derivative is always negative. 149 00:10:32,170 --> 00:10:34,990 And remember that the derivative represents the slope. 150 00:10:35,110 --> 00:10:40,840 So if the derivative is always negative the slope is always negative which means the series is always 151 00:10:40,840 --> 00:10:47,110 decreasing and if the series is always decreasing then a seven plus one is always going to be less than 152 00:10:47,430 --> 00:10:48,490 a 7. 153 00:10:48,490 --> 00:10:51,760 So based on that we can go ahead and check off these conditions. 154 00:10:51,760 --> 00:10:55,120 We know that both series are going to be greater than zero. 155 00:10:55,120 --> 00:10:59,740 They're always going to be positive but we also know that a seven plus one is always going to be less 156 00:10:59,740 --> 00:11:03,500 than a 7 because the series is decreasing. 157 00:11:03,580 --> 00:11:08,200 Our only condition left to check then is that the limit is and goes to infinity of the original series 158 00:11:08,200 --> 00:11:09,420 is equal to zero. 159 00:11:09,700 --> 00:11:15,100 So we just want to take the limit as and approaches infinity of our original series. 160 00:11:15,130 --> 00:11:19,250 Square root and plus one minus the square root of N.. 161 00:11:19,630 --> 00:11:24,610 If this limit is 0 then because we've already met this first condition will have met the second condition. 162 00:11:24,610 --> 00:11:29,590 Also we'll be able to prove that a 7 converges well just by looking at this we can see that the limit 163 00:11:30,010 --> 00:11:36,580 is going to tend towards zero because infinity plus one is not significantly different than infinity 164 00:11:36,760 --> 00:11:41,380 which means the square root of infinity plus one is not going to be significantly different than the 165 00:11:41,380 --> 00:11:42,700 square of infinity. 166 00:11:42,700 --> 00:11:48,740 Basically we're just taking the square of infinity minus the square of infinity which would be zero. 167 00:11:48,790 --> 00:11:50,320 So the limit will be zero. 168 00:11:50,320 --> 00:11:55,330 If you want to check that one thing you could do is multiply by the context here so we'd say the square 169 00:11:55,330 --> 00:12:00,160 root of and plus 1 instead of minus we do plus the square root of n. 170 00:12:00,430 --> 00:12:05,310 And then of course the denominator would be the same thing plus the squared of n. 171 00:12:05,380 --> 00:12:10,560 And if we did that then we would multiply out these two terms by these two terms in the numerator here 172 00:12:10,570 --> 00:12:17,650 so square root and plus 1 Time Square and plus 1 gives us and plus 1 we'd have squared and plus one 173 00:12:17,650 --> 00:12:22,240 time squared n but then minus squared and Times Square and plus 1. 174 00:12:22,240 --> 00:12:24,100 Those two terms that are on the net to zero. 175 00:12:24,340 --> 00:12:32,200 So then we just have minus route and times route n or minus N and then in the denominator we would end 176 00:12:32,200 --> 00:12:36,320 up with root and plus 1 plus root. 177 00:12:36,340 --> 00:12:40,770 And of course the limit as and approaches infinity. 178 00:12:40,780 --> 00:12:45,110 So then here you can see we're going to get an minus and to go to zero. 179 00:12:45,250 --> 00:12:52,810 So what we're left with then is just the limit as and goes to infinity of 1 divided by square root and 180 00:12:52,810 --> 00:12:55,690 plus 1 plus square root n. 181 00:12:55,960 --> 00:13:01,570 And here we can clearly see that if we were to evaluate as and goes to infinity we get infinity plus 182 00:13:01,590 --> 00:13:06,850 one or an extremely large number here taking the square root would stop a very large number plus the 183 00:13:06,850 --> 00:13:08,050 square root of infinity. 184 00:13:08,050 --> 00:13:09,490 Again a very large number. 185 00:13:09,490 --> 00:13:14,490 So what we basically end up with is 1 divided by infinity which is of course zero. 186 00:13:14,650 --> 00:13:19,910 So we can go ahead and say that this condition has been met as well since we've met all three conditions. 187 00:13:20,050 --> 00:13:26,500 We can say that by the alternating series Test this original series that we were given is going to converge 188 00:13:26,800 --> 00:13:32,020 and that's how you use the alternating series test to determine the convergence of an alternating series. 21217