- Worked example: convergent geometric series
- Worked example: divergent geometric series
- Infinite geometric series
- Infinite geometric series word problem: bouncing ball
- Infinite geometric series word problem: repeating decimal
- Proof of infinite geometric series formula
- Convergent & divergent geometric series (with manipulation)
Sal looks at examples of three infinite geometric series and determines if each of them converges or diverges. To do that, he needs to manipulate the expressions to find the common ratio. Created by Sal Khan.
Want to join the conversation?
- does a sequence have to converge if n approaches infinity if it's series converges or are they mutually exclusive?(6 votes)
- They can both converge or both diverge or the sequence can converge while the series diverge.
For example, the sequence as n→∞ of n^(1/n) converges to 1 . However, the series
∑ n=1 to ∞ n^(1/n) diverges toward infinity.
As far as I know, and I might be wrong about this (but I am fairly sure) that a sequence must converge in order for its series to converge (though the series is not certain to converge just because the sequence converges). So, yes, IF a series converges as n→∞, then its corresponding sequence must also converge.(5 votes)
- How do we find what the value converges to?(3 votes)
- @Arsenio Trucco note that that formula will only work for some rⁿ when n starts as zero. The sum from n=0 to infinity of a series is not always the same as the sum from n=5 to infinity of that series, because the first few terms are not counted towards the sum. You can compensate for this by using the proof in previous videos to discover that given that n starts at a constant b, Sn-rSn=ar^b, so Sn = (ar^b)/(1-r). This edited formula works for n=0 too, because ar^0 is just a, like the previous formula(4 votes)
- Could someone tell me what is the difference between a series and a sequence ?(2 votes)
- A sequence is a list of numbers in a specific order. A series is the result of adding the terms of a sequence.(5 votes)
- from3:47to3:56how did he move from 2^(n)/3^(n)3^(-1) to 3* 2^(n)/3^(n) ? Did he just multiply by 3 to cancel the 3^(-1) ?(4 votes)
- Well, 1 / (3^(-1)) = 1 / (1 / 3) = 1 * 3 / 1 = 3. So:
(2^n) / (3^n 3^(-1)) =
(2^n) / (3^n) * 1 / (3^(-1)) =
(2^n) / (3^n) * 1 / (1 / 3) =
(2^n) / (3^n) * 3 =
3(2^n) / (3^n)
Write it out on paper, without parentheses, and it may become clear.(2 votes)
- Around3:13, what was the mental math Sal did with the fractions? I understand that he was actually multiplying 1*3 and 2*9 and simplifying, but he muttered something about, "3 over 9... so we're gonna get 1." If he was using some kind of shorthand trick for multiplying fractions, that's a trick I'd love to know.(3 votes)
- I can't read his mind, but my guess would be: 3/2 * 1/9 = 3/9 * 1/2 = 1/3 * 1/2 = 1/6. In other words, he immediately saw that 3 and 9 are "related": 9 is divisible by 3, so if you have those on opposite sides of the fraction line (one above, one below) then you can simplify the fraction by dividing both by 3.(1 vote)
- if an infinite series is neither geometric nor arithmetic ,...e.g. 1+2x+3x^2+4x^3..... then please suggest some way to solve such problems(1 vote)
- Man I am so confused. Is there a reason he is using algebra to purposely arrive at only one n exponent?
For example, his first equation. (5^(n-1))(9/10)^n. Why can't we just immediately apply the limit as n approaches infinity?
(9/10)^n as n approaches infinity equals zero.
5^(n-1) as n approaches infinity equals infinity.
Infinity (zero) = converges at zero. But obviously that is not what he got. So confused.(1 vote)
- Remember that 0 and ∞ are approached, never equaled: so the rule that 0*anything = 0 does not apply when multiplied by ∞ because you have two rules in conflict. ∞ times anything approaches infinity while 0*anything approaches 0; thus these two rules conflict and the answer is indeterminate -- that is, the rules don't tell us what the limit is.(3 votes)
- Is it not easier to find r by simply dividing any a(n) by a(n-1)?(1 vote)
- You are referring to the ratio test. That will appear a little later on if you are working through chronologically.
- at2:04, why do you exclude 1/5 from the ratio? Doesn't it have to be (5*9)^n/(5*10)?(1 vote)
That's because he applies one of the laws of exponents where:
aˣ * bⁿ * cⁿ = aˣ * (bc)ⁿ.
It's possible to gather numbers or variables under a common exponent if they are raised to the same exponent. Since 1/5 is not raised to the nth power, it was excluded, yielding the value of 'a' in a * rⁿ. (the 10 in the denominator is raised to the nth power as well).(2 votes)
- He stated that it is an infinite series if 0<|r|<1 which is consistent with previous videos. At the end he stated it was a finite series when the constant was 2/3 and it was convergent. Did I miss something? Wouldn't that mean the convergent series was infinite because 2/3 is within the -1<r<1?
Which one is it exactly?
I just watched the geometric series videos in order and this one seems to really move a bit past the other videos. None of the previous videos had examples with anything other than k=0 under the sum and the video moved quickly past the calculations to the point that I would write them down and not really comprehend where the r value came from. I just know the final r value and know why he called it divergent or convergent, but even that made me miss whether it was going to be infinite or finite and confused me further when the statements didn't seem to match.(1 vote)
- No, if 0<|r|<1, that does not mean the series is infinite, it only means that it must converge if it is infinite. What makes a series infinite is the fact that you're taking it to an infinite number of terms (n=∞). You may be thinking of a finite sum (not a finite series), which is what a convergent series has.(2 votes)
- [Instructor] So here we have three different series. And what I would like you to do is pause this video, and think about whether each of them converges or diverges. All right, now let's work on this together. So, just as a refresher, converge means that even though you're summing up an infinite number of terms in all of these cases, if they converge, that means you actually get a finite value for that infinite sum, or that infinite number of terms being summed up, which I always find somewhat amazing. And diverging means that you're not going to get an actual finite value for the sum of all of the infinite terms. So how do we think about that? Well, we already know something about geometric series, and these look kind of like geometric series. So let's just remind ourselves what we already know. We know that a geometric series, the standard way of writing it is we're starting n equals, typical you'll often see n is equal to zero, but let's say we're starting at some constant. And then you're going to have, you're gonna go to infinity of a times r to the n, where r is our common ratio, we've talked about that in depth in other videos. We know, this is the standard way to write a geometric series. We know that if the absolute value of r is between zero, is between zero and one, then this thing is going to converge, converge. And if it doesn't, I'll just write it else, it will diverge. So maybe a good path would be, hey, can we rewrite these expressions that we're trying to take, that are defining each of our terms as we increment n, if we can rewrite it in this form, then we can identify the common ratio and think about whether it converges or diverges. So I'm going to focus on this part right over here. Let's see if I can rewrite that. So let's see, can I rewrite, let's see, five to the n minus one, I can rewrite that as five to the n times five to the negative one, and then that's gonna be times 9/10 to the n. And let's see, this is going to be equal to, going to be, I can just write this as this part right over here. I'll write it as 1/5, that's the same thing as five to the negative one times, and then five to the n and 9/10 to the n, well, I have the same exponent, so I can rewrite that as, and we're multiplying all this stuff, I'm just switching the order, this is the same thing as five times 9/10 to the nth power. And so this is going to be equal to 1/5 times, well five times nine is 45 divided by 10 is going to be 4.5, so times 4.5 to the nth power. So that original series I can rewrite as, just for good measure, I'm starting at n equals two, I'm going to infinity, and this can be rewritten as 1/5 times 4.5 to the n. So what's our common ratio, what's our r here? Well, you can see very clearly, it is 4.5. The absolute value of 4.5 is clearly not between zero and one. So this is a situation where we are going to diverge. Now if you found that inspiring, and if you weren't able to do it the first time I asked you to pause the video, try to pause the video again and try to work these out now, now that you've seen an example. All right, let's jump into it. So I'm just gonna try algebraically manipulate this part to get it into this form. So let's do that. So I can rewrite this, let's see, if I can get some things just to the nth power, so I can rewrite it as 3/2 to the nth power, and I could write this part right over here as times one over nine to the nth times nine squared. This is going to be equal to 3/2 to the n. And let's see, I could factor out or bring out the one over nine squared, so let me do that. So I'll write that as one over 81, I'll write it out there, so that's this part right over there. 1/81 times 3/2 to the n times one over nine to the n. But one over nine to the n, that is the same thing as one over nine all of that to the nth power. And the reason why I did that is now I have both of these things to the nth power, and I can do just what I did over here before. So this is all going to be equal to one over 81 times 3/2 times 1/9 to the nth power. These are just exponent properties that I am applying right over here. And so this is going to be equal to one over 81 times, let's see, 3/2 times 1/9 is 3/18 which is the same thing as 1/6, times 1/6 to the nth power. If I were to rewrite the original series, it's the sum from n equals five to infinity of, of, now I can rewrite it as one over 81 times 1/6 to the nth power. This is our common ratio, 1/6, very clearly. I'll do that in this light blue color, 1/6. That absolute value is clearly between zero and one, so this is a situation where we will converge, converge. Now let's do this last example. I'll do this one a little bit faster. So let's see, I could, if I'm just trying to algebraically manipulate that part there, that's going to be the same thing as two to the nth power times one over three to the n times three to the negative one. Well, this is going to be the same thing as two to the n times one over, why did I write equals? Times one over three to the n. And one over three to the negative one, that's the same thing as one over 1/3, that's just going to be equal to three times three. Well, that's going to be equal to, I'll give myself some space, we'll start out here. That's equal to, I'll put the three out front, three times two to the n times, and one over three to the n is the same thing as 1/3 to the nth power. And so this is going to be equal to three times two times 1/3, all that to the nth power. And so that's going to be equal to three times 2/3 to the nth power. So we just simplify this part right over here to three times 2/3 to the nth power. We can see that our common ratio is 2/3, so the absolute value of 2/3 is clearly between zero and one. So once again, we are going to converge, converge. And we're done.