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.
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.
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
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.
Fantastic as always man 🎉
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.
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.
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!
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
Nice!
Thanks!
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