Examples on Identifying Geometric Series and Determining if it Converges or Diverges by @mes on 3Speak
3Speak - YouTube - Telegram - Notes - Playlist - Sequences and Series - MES Links In this video, I go over two examples and two methods of identifying if a series is a geometric series, and then fi...
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Alright, so now let's look at an example that goes over the geometric series, just to find the sum of just a generic one. So here, find the sum of the geometric series, 5 minus 10 over 3 plus 20 over 9, minus 40, over 7, I mean, over 27 plus dot, dot, dot, notice how the signs change from negative, positive, negative. So let's look at the solution. So the first term is A equals 5, and the common ratio is negative. 2 over 3, and you can see this because, well, 5 is just A right there, so 5, that's just A, like that, and then we have minus 10 over 3, then plus 20 over 9, like that, I'll just write minus 40 over 27, plus dot, dot, dot. The difference between these two note right here is, I'll write this better. this is going to be well, negative 10 over 3 to get it to over here. We just have to multiply this by negative 2, 10 times 2 is 20, and then 3, so we have to have 3 like this, 3 times 3 is 9, which equals to 20 over 9. And then likewise from these 2 right here, 20 over 9 equals 2 times it by negative 2 over 3. This becomes negative 40 over, and then 3 times 9 is 27. So we get this one right there, and it keeps going on, same pattern. So hence, the common ratio is R there. Yes, so going further, so since R is negative 2 over 3, or the absolute value of R is equal to 2 over 3, and it's less than 1, so the value, just remove the positive, I'm making it positive, is less than 1, the series is convergent via example 1. So remember example 1, if it's between it, let's go to the summary of example 1. So if R is between or is less than absolute value of r is less than 1, then it's convergent, and its sums is A over 1 minus r. So, so this is convergent and its sum s equals to a minus 1 over r, A is equal to 5, and then R is 1 minus, and then we have and negative 2 over 3. So in other words, we have, we have like this, equals to 5, and then this is just going to be plus 2 over 3. So plus 2 over 3, and also multiply this by 3 or 3. So 1 is just 3 over 3, you get the common denominator, so we have equals to 5 over, this is going to be 5, and then 5 over 3, we'll just move the 3 on top, like that. just cancels equals to 3. Yeah, or for a completeness, just because I'll be doing this later in this video, I'll write this for completeness for this time, equals 5 over 3, and this equals 2, yeah, this just, the 3 goes on top. Yeah, 3 goes on top, in other words, we have a 1 over 1 over 3, like that. So yes, 1 divided by 1 divided by 3, the 3 goes on top. Yeah, just basic division, so this equals 3. Like that. Yeah, so a good question asks us, what do we really mean when we say that the sum of the series in example 2 is 3? Of course, we can't literally add an infinite number of terms, one by once, you can add them up, but according to definition 1, the total sum is the limit of the sequence of partial sums. So by taking the sum of sufficiently many terms, we can get as close as we like to the number 3. So the table shows the first-hand partial sums S-N, and the graph shows how the sequence of partial sums approaches three. Notice this one this says zig-zags across above and less than three, which is fascinating. So it doesn't need to be completely going towards three. In this case, it goes above it and below it. So here's a graph of it. So N is 1, 2, 3, all the way to 20, A.R, the power of n-minus-1, geometric series. where we have 5 like that, and then R is negative 2 over 3. And with Excel in that Excel sheet, it also have it in fractions, which is absolutely amazing. And then there's a summation right there. The number of the partial sum, this is a times 1 minus R and then divided by 1 minus R. So, but the limit approaches A over R. I mean, A over 1 minus R. is the limit there. Remember, that's just the partial sum. Scroll above to the partial sum before I take the limit. So right here, there's a partial sum over here. Going further, see this is fascinating stuff right here. And notice, yeah, start off at 5 and goes to here below 3, above 3, below 3, above 3, below 3, below 3, below 3. Notice here, 2.99, 2.99, and it goes up like this. Very, very fascinating graph like that. All right, so going further, we have now example 3, another example on geometric series. This one states, is the following series convergent or divergent, and it's here summation or sigma from n is 1, n equals to 1 to infinity, of 2 to the power of 2n, times 3 to the power of 1 minus n. So this is actually a geometric series, but it doesn't look like it initially, but what we could do is, well, let's look at the solution. Let's rewrite the n-term of the series in the form of ARN minus 1. We have to get it as a geometric series, if we can, and let's just see what we can come up with. So note that inside there, so we have the 2N. 2 to the power of 2 n times it by 3 to power of 1 minus n. What we could do is, well, let's take the 2 power out, so we can have this, same thing as writing 2 to the power of 2 to the power of n like that, times it by here to take a negative out, just factor negative out, so we'll have 3 times, well, negative, then becomes negative 1 plus n, so we have n minus 1, so we have this n minus 1 there. We're going further, so there's a negative there, and this one here is 2 to the power of 2, that's just equal to 4. So you have 4N, then get rid of this negative by making it at the bottom, so that it's 3 to the power of n minus 1. And now what I'll do is, because we want this R, the power of n minus 1, so basically there's only one power there is R, not A. So what I could do now is there's an n here, I could actually multiply this part right here, or yeah, multiply this by by 4, yeah, just by 4 times, yeah, 4 to the power of 1, times it by 4 to power of negative 1. In other words, this is just 4 divided by 4, like that. Yeah, doing so, we could just use the power laws. Right here, we can combine these two. Yeah, thus what we get is 4N minus 1 there, so it will have is equals to, equals to 4 n minus 1, so this combines with this one, and now what we'll do is we'll get rid of, oh, this one just goes over here, so times it by 4, like that, 4 to power of 1, then. Divide this by 3N minus 1. So now these are the like powers, you're dividing by, this 4 is put it in front, so we have 4, and then times it by, this is just going to be a fraction there. so the power can go on the outside 4 over 3, power of n minus 1. Yes, so thus what we have is put this all together with the summation, so we have summation of, or from n equals to 1 to infinity, equals to of this right here 2, the power of 2 n times it by 3, 1 minus n, equals to the summation as infinity. n equals to 1, there's 4 times 4 over 3, and minus 1. So in other words, we have 4 is A, and then this all 4 over 3 is R. And yeah, just write this as a geometric series, this equals to infinity like that, n equals to 1, a times it by, yeah, R, and minus 1, like that. And I'll just box this out like that. Yeah, so we recognize that this series as a geometric series with a equals 4 and r equals to 4 over 3, but since r is greater than 1, so r is 4 over 3, the series diverges as per example 1 of that summary of example 1, the geometric series. So only within, only when the absolute value of r is less than 1 is when it's going to be converging. So it diverges. And now note that another way to identify A and R is to write out the first few terms, and then just see the pattern there. So we have this formula like there, 2, the power of 2N. Yeah, so I'll just write this out like this. So we have infinity, then n equals to 1 of this right here, 2, 2n times it by 3, 1 minus n. So we just write out the terms, so for n equals to 1, we're going to have a 2, the power of 2 times it by 1 there, so 3, 1 minus 1, which is just 0, and this just goes to actually 1, 3 of the 0 is just 1, by definition. And then the next term 2, so plus 2 to the power of 2 times 2, that's just 2 to power of 4, like that, and then 3, and then times about 3, 1 minus 2 is negative 1. And then going further, we're going to have, for the third one, so 3 times 2 is 6, so 2 of the power of 6, and then 1 minus 3 is negative 2, so it's going to be 3, and then negative 2, and then we keep going on and on, and plus, and I'll just show you the pattern that we have right here, this one's going to equal to 4, I guess 4 times 1, plus here 2 times, yeah, 2 the power of 4, This equals 2, well, 2 times 2 is 4, times that by 2 is 8, and then times it by another 2, so we have 4 2 is 16. So 2 to the power of 4 is 16, divided by 3 times it times 3 to the power of negative 1, that's just divided by 3. And then over here is 6 right there. So we have to multiply this 2 to the 4 by 2 twice. So we get 2 times about 16 times 2 again, 64. So we're going to have a plus. 64, then divided by here 3 to power of 2, which is just 9, like that. So that's the pattern plus dot, dot, dot, keeps going on and on. And we just want to look at the first few terms. You know, it's the geometric series. A is equal to 4, and then notice the difference now here, so that's A. So this one here, we have, well, 16 over 3 times it by here is... This one is 16 times 2 is 32 times 2 is 64, so in other words, times it by 4 to get 64, and then 3 times 3 to get 9. So this equals to 64 over 9. So in other words, R is 4 over 3. So yes, yeah, multiple ways of getting the A and the R for the geometric series.