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.
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.
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.
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.
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.
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.
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
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.
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).
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/n.gif)
You left off 0/1, which makes it easy.
What about -1/0? Then we get to the real fun part
@@arthurspuri2948 Or how about defining a function f(x) = (x / (1+x)) * ((1+x)/(2+x))*...?
@@arthurspuri2948then the zeros cancel, ez 🤓👍
(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
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.
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.
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.
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.
Fantastic as always man 🎉
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.
3/4*8/9*15/16*24/25*35/36*…=0.5=1/2
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.
Nice. This is simplest and clearest imo.
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.
Good point!
Nice!
Thanks!
Wolfram Alpha prompt:
product[ (n^2-1)/(n^2), {n,2,∞}]
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
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.
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).
calm down
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
Thaty the way i did it to.
@@davidbelgardt3775❤
@@davidbelgardt3775👍
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