An Interesting Infinite Product

[@christosvoskresye]

You left off 0/1, which makes it easy.

[@arthurspuri2948]

What about -1/0? Then we get to the real fun part

[@christosvoskresye]

@@arthurspuri2948 Or how about defining a function f(x) = (x / (1+x)) * ((1+x)/(2+x))*...?

[@alex4455]

@@arthurspuri2948then the zeros cancel, ez 🤓👍

[@TheMathManProfundities]

Then it'd just be 0.

[@christosvoskresye]

@@TheMathManProfundities How long did it take you to work that out?

[@Qermaq]

Interestingly, the product of (n^2 - 1)/(n^2) from n = 2 to k works out to be (k+1)/(2k). From this, it's simple to see that as k tends to infinity the +1 fades away and we get (k)/(2k) or 1/2.

[@SyberMath]

Good point!

[@rainerzufall42]

1:27 I can feel the pattern (1 + 1/(n+1)) / 2 = (n + 2) / (2 n + 2) converging to 1 / 2 ... let's see!

[@senshtatulo]

The product symbol is capital pi, Π, not just a larger lowercase pi, π.

[@RooiGevaar19]

In Greek calligraphy both uppercase and lowercase pis are the same in shape. Same in Cyrillic.

[@senshtatulo]

@@RooiGevaar19 Are you suggesting that this math usage is calligraphic?

[@RooiGevaar19]

@@senshtatulo yes. It's size that matters whether it's lower case or upper one. Same way א (aleph) may be written in its calligraphic form, as the printed one might be problematic for some people - however most people don't speak Hebrew and don't know how to write א, and they invent their own awful cursives (the actual cursive of א looks like a disjoint K, or like "|C").

[@senshtatulo]

@@RooiGevaar19 I don't see your point. The example with aleph seems beside the point. Could you explain in other words?

[@pierreabbat6157]

8:30 You can't do that! Π((n+1)/n) diverges, like the harmonic series. You get 0×∞. Btw it's Π, not π.

[@nerd8502]

You can do that for the same reason you can get the limit of 1/x * x as x approaches infinity. Sure, when considered individually you end up with 0 times infinity, but by considering the whole formula as a single thing, you simplify to x/x and get a limit of 1

[@swthiel]

Once you've written the product as P=Π[(n+1)(n-1)/n^2 (n= 2, 3...), it's not too tricky to rewrite the product as P=Π(n+1)Π(n-1)/P(n^2) with (n= 2, 3...) and then re-index the (n+1) and (n-1) products to get everything to cancel within a big product (n = 3, 4, ...) except for some leading terms that give you the 1/2.

[@vcvartak7111]

Splitting the two infinite products is nice technique

[@SyberMath]

It sure is! 😍

[@simmmr.9040]

if we represent each fraction as (n²-1)/n² and then open the brackets, we get (n-1)(n+1)/n², then the first multiplier of the numerator will be reduced with the previous denominator, and the second multiplier of the numerator with the next denominator - finally we come to the product of fractions (2-1)/2 and (n+1)/n, which is equal to 1/2 at n→∞

[@farhansadik5423]

Fantastic as always man 🎉

[@SyberMath]

Thank you, sir! 😍

[@MathNotationsVids]

(n-1)/n telescopes to 1/n (n+1)/n telescopes to (n+1)/2 Therefore, the partial products are of the form (1/2) • ((n+1)/n) which --> 1/2

[@chaosredefined3834]

You have to be careful with telescoping series. It's not a problem here, but... the infinite sum is a limit. So, what you have is (1/2)(2/3)...((N-1)/N) (3/2)(4/3)(5/4)...((N+1)/N). This simplifies down to (N+1)/2N, which does converge to 1/2. Black pen red pen did take a series where the telescoping limit didn't go as nicely, because the final term never completely disappeared. Can't remember the exact sequence unfortunately, but you do need to double check that the limit of the telescoping cancellations is 1.

[@AbdelrahmanYasser-tt3yi]

pls do things like that a lot because i want to prove very hard formula. talking about these products

[@discreaminant]

The real difficulty for me is to even figure out the pattern lmao 🗿

[@SyberMath]

Might take some practice. Two things to notice: 1- The denominators are one more than the numerators 2- The denominators are perfect squares

[@Blaqjaqshellaq]

3/4=1/2 + 1/4 (3/4)*(8/9)=2/3=1/2 + 1/6 (3/4)*(8/9)*(15/16)=5/8=1/2 + 1/8 (3/4)*(8/9)*(15/16)*(24/25)=3/5=1/2 + 1/10 See the pattern? Each product is an entry in the series (1/2)*(1+1/n), which converges at 1/2.

[@dantallman5345]

Nice. This is simplest and clearest imo.

[@NightSkyJeff]

8:30 isn’t the left product expansion erroneous? The numerator is supposed to be smaller than the denominator.

[@jursamaj]

Did you watch past that point? He corrected himself.

[@NightSkyJeff]

@ yes I did.

[@samueljehanno]

@@jursamaj yes I did... That's not rigorous

[@Xavras_Wyzryn98]

It's a product of (x-1)(x+1)/x^2 from 2 to infinity, in any finite version of this product all terms beside 1st (x-1)/x and last (x+1)/x cancel out. With limes x--> +inf it comes down to (1/2)*lim x--> +inf (x+1)/x Which is 1/2*1, so 1/2 is the answer.

[@someperson188]

The nth multiplicand is n(n + 2)/(n + 1)^2, for n = 1, 2, 3, ... Lemma. The nth partial product is (n+2)/(2(n+1)). Proof. The Lemma holds for n = 1. Suppose the Lemma holds for some n. The (n+1)th multiplicand is (n+1)(n+3)/(n+2)^2. Thus, the (n+1)th partial product is [(n+2)/(2(n+1))][(n+1)(n+3)/(n+2)^2] = (n+3)/(2(n+2)). proving the Lemma by induction. The limit of the partial products is now clearly 1/2.

[@dan-florinchereches4892]

I would say the for of the product is π (i^2-1)/i^2=π (i-1)(I+1)/I^2 For I=2 up to N for each product i-1 in numerator will simplify with an i-1 in the previous denominator and I+1 with an I+1 in the next denominator so by simplifying π=1/2*(n+1)/N

[@ignaciodecastrofondevila2456]

In 07:37 you state that Product[f(n)*g(n)]=Product[f(n)]*Product[g(n)], but this is only true on the condition that such products exist, which you have not proved previusly. In this case, proving the existence of Product(n-1/n) is easy (factors less than 1), but proving the existence of Product(n+1/n) is not so easy. Thank you very much for your video, wonderful as always.

[@SyberMath]

Thank you! I barely do proofs. I just assume 😄😁

[@dixonblog]

Wolfram Alpha prompt: product[ (n^2-1)/(n^2), {n,2,∞}]

[@Qbe_Root]

You can also keep going with Π notation to make it more rigorous: Π(n = 2, ∞, (n - 1) / n) = Π(n = 1, ∞, ((n + 1) - 1) / (n + 1)) = Π(n = 1, ∞, n / (n + 1)) = 1 / (1 + 1) × Π(n = 2, ∞, n / (n + 1)) so Π(n = 2, ∞, (n - 1) / n) × Π(n = 2, ∞, (n + 1) / n) = 1 / 2 × Π(n = 2, ∞, n / (n + 1)) × Π(n = 2, ∞, (n + 1) / n) = 1 / 2 × Π(n = 2, ∞, n / (n + 1) × (n + 1) / n) = 1 / 2 × Π(n = 2, ∞, 1) = 1 / 2 × 1 = 1 / 2

[@SamanNagme]

Can you please tell us what software you use, from mobile phone or pc? Thanks

[@SyberMath]

chk description

[@SamanNagme]

@ thanks for your reply I know it’s you solving the math problems not the software but I want to make a video to type math equations and record it if you can help please. Thanks

General solution. Reindex so you start from 1. Product of (n(n+2))/((n+1)(n+1)), with solution 1!1!/0!2! = 1/2. (Obtained by the factorials of the numbers after n in the denominator placed in numerator and the factorials of the numbers after n in numerator placed in the denominator. (Negatives of the roots of the polynomials).That which determines the partial products are in the usual order). Partial products are (1/2) n!(n+2)!/((n+1)!(n+1)!). So that the partial products for the product in the video starting at 2 is (1/2) (n-1)!(n+1)!/(n!•n!). (Works also for non integers using the continuous factorial.).

[@MisterFourSeventeen]

7:59 Without first checking whether the product converges absolutely (by checking the exponential of the infinite sum of the natural logarithmic terms), that is a dangerous move.

[@MisterFourSeventeen]

I did an integral test using S (2~inf.) |ln(1-1/x^2)|dx with a substitution of v = 1-1/x^2 (thus changing the bounds to 0.75 ~ 1) to convince myself of the convergence of the infinite series of ln(1 - 1/n^2) starting at n = 2.

[@SyberMath]

it converges. it's obvious 😄

[@MisterFourSeventeen]

@@SyberMath I wouldn't call it obvious when limit test (which result in 0) and ratio test (which result in 1) are inconclusive. Sometimes what seems obvious may not be so obvious after all.

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@penfelyn]

so why is the first attempt incorrect? what is wrong with that kind of reasoning?

[@SyberMath]

I think I made a mistake

[@russellsharpe288]

The partial products of Π((n+1)/n) starting at n = 2 are 3/2, (3/2).(4/3) = 2, (3/2).(4/3).(5/4) = 5/2, (3/2).(4/3).(5/4).(6/5) = 3 and in general the partial product up to n = N is (N+1)/2. Clearly this diverges as N -> infinity, and is certainly not equal to 1/2 as you seem to think. Similarly the partial products of Π(n/(n+1)) are 1/2, (1/2)(2/3) = 1/3, (1/2)(2/3)(3/4) = 1/4 and the infinite product converges to zero. So your calculation at the end becomes zero x infinity: undefined. I daresay you cannot rearrange terms of an infinite product as you have here unless the corresponding infinite sum (taking logarithms termwise) is absolutely convergent (Riemann Rearrangement Theorem).

[@giuseppemalaguti435]

S=Π(k^2-1)/k^2..(k=2,3...)..applico ln .lnS=Σln((k^2-1)/k^2)=Σln(k^2-1)-Σlnk^2=Σln(k+1)+Σln(k-1)-2Σlnk=(ln3+ln1-2ln2)+(ln4+ln2-2ln3)+(ln5+ln3-2ln4)+(ln6+ln4-2ln5)....=ln1-ln2=-ln2=ln(1/2)...S=1/2

[@davidbelgardt3775]

Thaty the way i did it to.

[@giuseppemalaguti435]

​@@davidbelgardt3775❤

[@giuseppemalaguti435]

​@@davidbelgardt3775👍

[@faeancestor]

calm down

[@SyberMath]

ok

[@hrk992]

prod[(k²-1)/k²]; k=2...n = prod[(k²-1)]/(n!²); k=2...n = prod[(k-1)(k+1)]/(n!²); k=2...n = (n-1)!(n+1)!/(2n!²) = 1/2+1/(2n) lim[1/2+1/(2n)]; n → oo = 1/2

[@yhira2010]

The denominator is (n+i)^2 the numerator is [(n+i)^2-1]=[(n+i+1)(n+i-1)] i=0,1,2,3,,,,,♾️ n=2 so the final form of this multiplization get at [(n-1)/n] ×[(♾️+1)/♾️] (*) =[1/2]×1 =1/2 (*)cause of mutual cancellation in the consecutive three numerators and denominators