Algebra (all content)
Sal evaluates the infinite geometric series 8+8/3+8/9+... Because the common ratio's absolute value is less than 1, the series converges to a finite number.
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- Where can I find the video that derives the formula for the sum of infinite geometric series?(55 votes)
- I see you were given a link. The proof is very simple:
𝑆 = 𝑎₁ + 𝑎₁𝑟 + 𝑎₁𝑟² + ...
𝑆 = 𝑎₁ + 𝑟(𝑎₁ + 𝑎₁𝑟 + ...)
𝑆 = 𝑎₁ + 𝑟𝑆
𝑆 – 𝑟𝑆 = 𝑎₁
𝑆(1 – 𝑟) = 𝑎₁
𝑆 = 𝑎₁/(1 – 𝑟)(138 votes)
- Link to video on infinite geometric series:
- Can sigma notation and S sub n be used interchangeably?(6 votes)
- No, sigma notation is something TOTALLY different. S sub n is the symbol for a series, like the sum of a a geometric sequence. Sigma notation is just a symbol to represent summation notation. Honestly, I see what you mean, about them seeming the same, but I don't think it could be used interchangeably.(3 votes)
- Would anyone happen to know if there any videos on arithmetic sequences?(4 votes)
- Am I visualizing this correctly?
So let's suppose we have a piece of bread that is 12 cm long. For the first slice, I cut it at exactly 8 cm and the slices that follow are one-third the length of the previous slice, does that mean that I will never run out of bread to divide given that I have the ability to precisely slice it at the correct length?(3 votes)
- In the situation you describe, the lengths can be represented by the 8 times the geometric series with a common ratio of 1/3. The geometric series will converge to 1/(1-(1/3)) = 1/(2/3) = 3/2. You will end up cutting a total length of 8*3/2 = 12 cm of bread. So, you will never run out of bread if your first slice is 8cm and each subsequent slice is 1/3 as thick as the previous slice.(3 votes)
- Why is the limit of (1/3)^k at k = infinity not 0?(2 votes)
- I am thinking, say r=1 then for a rational fraction we would get a domain issue right? a/0. Where you get a vertical asymptote. In this case even if it follows a formula that comes from a geometric series I am guessing we also view it in the same way, actually there is a domain issue when r=1 then basically you have an infinite sum of the the first term. Your output is infinity and you get an asymptote. Is my understanding correct? I am trying to visualise this graphically... not sure if my thinking is correct or not.(2 votes)
- A convergent geometric series is such that the sum of all the term after the nth term is 3 times the nth term.Find the common ratio of the progression given that the first term of the progression is a. Show that the sum to infinity is 4a and find in terms of a the geometric mean of the first and sixth term(1 vote)
- at2:45, so when the geometric series diverges (|r|≥1), we cannot use the formula a/(1-r)?(1 vote)
- Correct. The idea of a sum breaks down for a divergent series, as the terms don't converge to zero and hence, a sum cannot exist. So, when a geometric series diverges, that's all there is to it. You can't really do much more with it.(2 votes)
- [Voiceover] Let's get some practice taking sums of infinite geometric series. So we have one over here. And just to make sure that we're dealing with a geometric series, let's make sure we have a common ratio. So let's see, to go from the first term to the second term we multiply by 1/3, then go to the next term we are going to multiply by 1/3 again, and we're going to keep doing that. So we could rewrite the series as eight plus eight times 1/3, eight times 1/3, plus eight times 1/3 squared, eight times 1/3 squared. Each successive term we multiply by 1/3 again. And so when you look at it this way, you're like, okay, we could write this in sigma notation. This is going be equal to, so this, the first thing we wrote is equal to this, which is equal to, this is equal to the sum. And we could start at zero or at one, depending on how we like to do it. We could say from k is equal to zero. And this is an infinite series right here, we're just gonna keep on going forever, so to infinity of, well what's our first term? Our first term is eight. So it's going to be eight times our common ratio, times our common ratio 1/3 to the k power. Now, let me just verify that this indeed works. And I always do this just as a reality check, and I encourage you to do the same. So, when k equals zero, that should be the first term right over here. You get eight times 1/3 to the zero power, which is indeed eight. When k is equal to one, that's gonna be our second term here. That's gonna be eight times 1/3 to the first power. That's what we have here. And so when k is equal to two, that is this term right over here. So these are all describing the same thing. So now that we've seen that we can write a geometric series in multiple ways, let's find the sum. Well, we've seen before and we prove it in other videos, if you have a sum from k equal zero to infinity and you have your first term a times r to the k power, r to the k power, assuming this converges, so, assuming that the absolute value of your common ratio is less than one, this is what needs to be true for convergence, this is going to be equal to, this is going to be equal to our first term which is a over one minus our common ratio, one minus our common ratio. And if this looks unfamiliar to you, I encourage you to watch the video where we find the formula, we derive the formula for the sum of an infinite geometric series. But just applying that over here, we are going to get, we are going to get, this is going to be equal to our first term which is eight, so that is eight over one minus, one minus our common ratio, over 1/3. And we know this is going to converge, because our common ratio, the magnitude, the absolute value of 1/3 is indeed less than one. And so this is all going to converge to, this is going to converge to eight over one minus 1/3 is 2/3, which is the same thing as eight times 3/2, which is, let's see this could become, divide eight by two, that becomes four, and so this will become 12.