Log in Paze 11 years agoPosted 11 years ago. Direct link to Paze's post “There's something wrong w...” There's something wrong with my calculations; somebody please help me. If we take the ratio to be 2, then the result of the sum would be +infinite. X = 5 + 5*2 + 5*2² + 5* 2³ etc.... What's wrong with this logic? • (24 votes) Ethan Dlugie 11 years agoPosted 11 years ago. Direct link to Ethan Dlugie's post “Sal said it himself at th...” Sal said it himself at the end, that the common ratio _r_ must satisfy the condition |r|<1. This is not the case for your specific sum. Dealing with infinity is in general a dangerous venture and can get you into a lot of trouble if you don't treat it vigorously. (33 votes) adamscarlat 9 years agoPosted 9 years ago. Direct link to adamscarlat's post “In the derivation of the ...” In the derivation of the finite geometric series formula we took into account the last term when we subtracted Sn-rSn and were left with a-ar^(n+1) in the numerator. Here Sal subtracted Sinf-rSinf and sort of ignored the last term and just had the numerator to equal a. • (17 votes) Stefen 9 years agoPosted 9 years ago. Direct link to Stefen's post “Ooooh - the mysteries of ...” Ooooh - the mysteries of infinity! The thing of it is, THERE IS NO LAST TERM. That means, with respect to Sn and rSn, all terms cancel out except the first term in Sn, a_0 since for all other terms Sn has just as many as rSn. (15 votes) Jai Sankar 10 years agoPosted 10 years ago. Direct link to Jai Sankar's post “At 7:27, Sal said that if...” At • (5 votes) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “Your logic seems plausibl...” Your logic seems plausible but fails the epsilon-delta test, which is the ultimate test for whether a series converges. There is no delta for which larger values of n will produce S values closer than, say, a/4 (a possible epsilon). We know this because it's clear that there is no point beyond which the sum stops oscillating between a and 0. Your result of a/2 is the average of the two values between which the sum oscillates, not the value to which the sum converges. (14 votes) bb 11 years agoPosted 11 years ago. Direct link to bb's post “Please, help me understan...” Please, help me understand the parts in this proof: • (6 votes) Charles Russell 11 years agoPosted 11 years ago. Direct link to Charles Russell's post “If you divided a by r the...” If you divided a by r the first term would become a/r which is no good. By multiplying both sides by r you create a kind of phase shift of terms such that by subtracting like terms from both series you are left with a first term and a last term that goes on to give the required sum. (6 votes) Abdullah 9 years agoPosted 9 years ago. Direct link to Abdullah's post “I think you forgot to put...” I think you forgot to put ar^(n+1) in the second series • (4 votes) 9 years agoPosted 9 years ago. Direct link to Stefen's post “We don't need to since bo...” We don't need to since both series have an ar^(n+1) term, which cancel out, leaving only the ar^0 term. If that seems weird to you, and if this is your first exposure to sequences/series that go to infinity, then the mathematical concept of infinity can seem a bit strange. (8 votes) Rainman 9 years agoPosted 9 years ago. Direct link to Rainman's post “what will be the sum of i...” what will be the sum of infinite geometric series 2/3 + 1/3 + 1/6..... up to 8 term? and what does that mean up to 8 term? • (2 votes) Thomas B 9 years agoPosted 9 years ago. Direct link to Thomas B's post “Up to the 8th term would ...” Up to the 8th term would imply that you are only summing the first 8 terms, not the entire series. (3 votes) Lily G. 9 years agoPosted 9 years ago. Direct link to Lily G.'s post “So, this particular formu...” So, this particular formula would work if "r" were anywhere between -1 and 0, and 0 and 1, right? • (5 votes) Stefen 9 years agoPosted 9 years ago. Direct link to Stefen's post “That is correct. If the a...” That is correct. If the absolute value of r is 1 or greater, that is if r IS NOT -1<r<0 or 0<r<1, (as you observe) then the sum diverges (goes to infinity) and the formula will not give a correct answer. (3 votes) Dominic Chan 10 years agoPosted 10 years ago. Direct link to Dominic Chan's post “what if u times r^2 to S(...” what if u times r^2 to S(n) then won't you get S(infinity) - r^2 S(infinity) = to ar^(0) + ar^(1) • (2 votes) doctorfoxphd 10 years agoPosted 10 years ago. Direct link to doctorfoxphd's post “Well, let's try that: S...” Well, let's try that: (7 votes) Coco 9 years agoPosted 9 years ago. Direct link to Coco's post “In order to get all terms...” In order to get all terms to cancel except for a, we must assume that Sinfinity has one extra term than r*S infinity. Someone said there is no last term in an infinite series. But I am not talking about a last term, just about the fact that Sinfinity must have one extra term (if there isn't the cancellation does not work). Since r*Sinfinity has one less term, it is finite (or has an upper bound). So is r, if <1, an operator on infinite series, bounding them, or making them finite? • (2 votes) Stefen 9 years agoPosted 9 years ago. Direct link to Stefen's post “Infinity is not a number,...” Infinity is not a number, it is a concept. S and rS have the same number of terms. The difference is that S has a term that rS does not. What happens is when you get way way way out there, that is, the value of n for S_n and rS_n is very very very large, the value of each term of S_n and rS_n become negligible. If you do not think each term is small enough to be accurate for your needs, then keep on increasing the size of n until it does satisfy your needs. The next weird thing is that if you let n go to infinity, the sequences n and 2n have the same number of terms! I highly recommend that you spend some time in personal research on the concept of infinity. Good luck and keep studying! (6 votes) christophorino 9 years agoPosted 9 years ago. Direct link to christophorino's post “At the end, Sal said that...” At the end, Sal said that 5/2/5 is 12.5. But isn't it one half? 5/2*5, 5/10, 1/2 • (2 votes) redthumb.liberty 9 years agoPosted 9 years ago. Direct link to redthumb.liberty's post “You need to be very caref...” You need to be very careful with multiple divisions or divisions of fractions. I always make it a point to model those types of fraction divisions as: That helps to keep me from falling into the invalid reduction trap that you just fell into. (5 votes)Want to join the conversation?
But let's put it in numbers in the same way Sal did:
now we multiply X by r, which is 2, and then let's subtract them.
Now, X-2X = 5
X=5/1-2
X=-5 (!)
It should be +infinite, right?
Here is a simple yet interesting example I found on wikipedia:
∑ 0 from n=1...oo (oo denotes infinity)
This sum is clearly 0, but we can do a little math trickery...
=∑ (1-1) from n=1...oo
=(1-1)+(1-1)+...=1+(-1+1)+(-1+1)+...=
1+∑(-1+1) from n=1...oo
=1+∑0
=1
Which is definitely not right.
My question is, in the case of the infinite series, how can you rigorously prove that every term does cancel out?
Thanks!
Part of the formality you may be missing is noting that as n goes to infinity, the limit of each term a_n which is composed of (a_0)r^n goes to zero. Here is another way of writing the proof that uses the limit argument to get to the same place Sal did: http://bajasound.com/khan/khan0009.jpg.
To understand the proof, you need to understand that for |r|<1 the limit of r^n as n-->infinity is zero.
Great Question!
Keep Studying!
7:27
why do we multiply and after that, we subtract the first sum from the second sum, because these two steps give us the formula, but based on what reasoning we do this exact step: first multiply by the common ratio and second - subtract the second sum from the first sum?
thanks
Try this video to get another perspective on the concept of infinity:
https://www.youtube.com/watch?v=elvOZm0d4H0
and S(infinity) = (ar^(0) + ar^ ( 1 ) ) / (1 - r^2)
Starting with
S∞ = ar⁰ + ar¹ + ar² + ar³ + ar⁴ + ar⁵ + ar⁶ + ar⁷ . . .
multiply that series by r²
r²∙S∞ = r²∙ar⁰ + r²∙ar¹ + r²∙ar² + r²∙ar³ + r²∙ar⁴ + r²∙ar⁵ + . . .
Simplify
r²S∞ = ar² + ar³ + ar⁴ + ar⁵ + ar⁶ + ar⁷ + . . .
Then subtract
S∞ - r²S∞
S∞ = ar⁰ + ar¹ + ar² + ar³ + ar⁴ + ar⁵ + ar⁶ + ar⁷
- r²S∞ = - ar² - ar³ - ar⁴ - ar⁵ - ar⁶ - ar⁷
Resulting in
S∞ - r²S∞ = ar⁰ + ar¹ which is equivalent to a + ar
S∞(1 - r²) = a + ar
and
S∞ = (a + ar)/(1 - r²) your result
That isn't simplified, though:
You can factor further, I guess
S∞ = a(1 + r)/[(1 + r)(1 - r)]
Cancel the like factors of (1 + r)
Back to S∞ = a/(1 - r)
Here are some links to get you started:
http://www.emis.de/proceedings/PME30/4/345.pdf - this is about typical ways infinity is understood by students
https://www.youtube.com/watch?v=elvOZm0d4H0 - this is about the various types of infinities
Also 5/5 is 1, 1 halved is 1/25/2/5 is actually 5/1 ÷ 2/5, which is evaluated: 5/1 * 5/2 = 25/2 = 12.55/(2/5) → 5*(5/2)
Video transcript
What I want to do isanother "proofy-like" thing to think about the sum ofan infinite geometric series. And we'll use avery similar idea to what we used to find the sumof a finite geometric series. So let's say I have a geometricseries, an infinite geometric series. So we're going tostart at k equals 0, and we're never going to stop. We're it's going allthe way to infinity. So we're never going tostop adding terms here. And it's going to be our firstterm times our common ratio. Our common ratioto the kth power. Actually let me dok and that color. k equals 0 all theway to infinity. And so let's just callthis thing right over here, let's call this s sub infinity. We're going all the way toinfinity right over here. And so this, if wewere to expand it out is going to be equal to atimes r to the 0-- actually let me just write it outlike that which is just a. a times r to the 0 power plusa times r to the 1st power. r to the 1st power. Plus a times r to the 2nd power. r to the 2nd power. Plus-- and we could justkeep going on and on and on. I think you getthe general idea. Now just like whenwe tried to derive a formula for the sum ofa finite geometric series we just said, well whathappens if you take the sum and if you were tomultiply every term by your common ratio. Every term by r. So let's do that. Let's imagine this sum. And we're going tomultiply every term by r. And the reasonwhy I said this is "proofy" is this is not alwaysclear-- It's a little bit, when you're multiplyingsomething times infinite terms or an infinite sum, atleast this will be at least give you the general idea. Or when you start thinkingabout and infinity, sometimes I just think aboutthings a little bit deeper. So r times this infinite sum? Well that's going tobe equal to-- We're just going to multiplyevery term here times r. So a r to the 0'th power timesr is going to be a times r. a times r to the 1st power. Multiply this one times r? You're going to get atimes r to the 2nd power. a times r to the 2nd power. I think you seewhere this is going. Multiply this one times r? You're going to get plus atimes r to the 3rd power. And we would just keep on going. We'd just keep on going. So let me just show that. So plus dot dot dot. Now what happens ifwe were to subtract this sum from this top sum? So on the lefthand side, we could express that as our sum s subinfinity minus our common ratio times s sub infinity. Is going to be equal to-- So when you subtract you'regoing to have a times r to the 0'th power, which isreally just the same thing as a. That's just going to bea. a times r to the 0 is just a times 1 which is a. We'll write in that same color. Is equal to a. But every otherterm, you're going to have a times r tothe 1st, but you're gonna subtract atimes r to the 1st. You have a times r tothe 2nd, but you're going to subtract atimes r to the 2nd. So every other term isgoing to be subtracted away. And this happens allthe way to infinity. It never, never ends. So the only termthat you're left with is just thatfirst one, is just a. And so now we can actuallytry to solve for our sum. If you factor outthe s sub infinity, you are left with 1 minus r. 1 minus r. s times s, our sum, times1 minus r is equal to a. Divide both sidesby 1 minus r, and we get that our sum, thething that we cared about-- And once again, this iskind of an amazing result. That we're taking the sum ofan infinite number of terms and under theproper constraints, we are going toget a finite value. So this is going to beequal to a over 1 minus r. So once again,it's kind of neat. If let's say I had the sum,let's say we started with 5, and then each time wewere to multiply by 3/5. So 5 plus 3/5 times 5 is 3,times 3/5 is going to be 9/5. 9/5, or I'll multiply by 9/5again-- Oh, sorry not 9/5. My brain isn't working right! 5 times 3/5 is goingto be 3 times 3/5. Is going to be-- 3times this is going to be 9/5-- actuallythat was right. My brain is working right. Times 3/5 is goingto be 27 over 25. Times 3/5 is going to be 81/125. And we keep on going onand on and on forever. And notice theseterms are starting to get smaller andsmaller and smaller. Well actually allof them are getting smaller and smaller and smaller. We're multiplyingby 3/5 every time. We now know what thesum is going to be. It's going to be ourfirst term-- it's going to be 5-- over 1minus our common ratio. And our common ratioin this case is 3/5. So this is going to beequal to 5 over 2/5, which is the samething as 5 times 5/2 which is 25/2 which is equalto 12 and 1/2, or 12.5. Once again, amazing result. I'm taking a sum ofinfinite terms here, and I was able toget a finite result. And once again, whendoes this happen? Well, if our common ratio--if the absolute value of our common ratio--is less than 1, then these termsare going to get smaller and smaller and smaller. And you'll even see here iteven works out mathematically in this denominatorthat you are going to get a reasonable answer. And it makes sensebecause these terms are getting smallerand smaller and smaller that this thing will converge. Even if r is 0. If r is 0, we're stillnot dealing strictly with a geometric series anymore,but obviously if r was 0, then you're really only goingto have this-- well, even this first term iskind of under debate depending on how youdefine what 0 to 0 is. But if your first termyou just said would be a, then clearly you'd just be leftwith a is the sum, and a over 1 minus 0 is still a. So this formula that we justderived does hold up for that. It does start to break down ifr is equal to 1 or negative 1. If r is equal to 1 thenas you imagine here, you just have a plusa plus a plus a, going on and on forever. If r is equal to negative 1you just keep oscillating. a, minus a, plus a, minus a. And so the sum's value keepsoscillating between two values. So in general thisinfinite geometric series is going to convergeif the absolute value of your common ratiois less than 1. Or another way of sayingthat, if your common ratio is between 1 and negative 1.
