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anonymous
 3 years ago
Can someone PROVE this:
anonymous
 3 years ago
Can someone PROVE this:

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358686236762:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is dw:1358696599099:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0oh, \[^{2n2}C_{n1}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1358757323764:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358757562194:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2I think this is untrue, the LHS diverges while RHS converges

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How can we say LHS diverges?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2the LHS is of order n, definitely it diverges

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0RHS is convergent as per p test

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2\[ \frac{2 (n1)^2}{8n} = \frac{n^2  2n + 1}{4n} = \left( \frac{n}{4}  \frac{1}{2} \right) + \frac{1}{4n}\] Both part of above diverges.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The actual question is here http://openstudy.com/users/jonask#/updates/50ed3bece4b0d4a537cd13df

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LHS is divergent by comparison test

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358758646393:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2let me check this sequence with mathematica

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what result did u get from mathematica?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2So fat the result is positive for that sequence.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0up to how many terms u tried?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2for 1000 the result is 0.982152

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2The compact form of the recurrence relation is \[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n1)}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2Here's a mathematica code for it Table[(2 n)!/((n!)^2 2^(2 n) (2 n  1)), {n, 1, 10}] and to verify it a = 1/2 For[i = 2, i <= 10, i++, a = (2 i  3)/(2 i) a; Print[a];] This converges exactly to 1 Sum[Binomial[2 n, n]/(2^(2 n) (2 n  1)), {n, 1, Infinity}]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2\[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n1)} = \frac{1}{2^{2n}(2n1)} \times \binom{2n}{n} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could somebody give a medal please

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If u will prove it may get more than 1 medal

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am only I 6 grade I havent done anything like that before I am sorry. but could you please give me a medal

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2you can show that \[ \sum_{n=1}^\infty \frac{1}{2^{2n}(2n1)} \times \binom{2n}{n} = 1 \] using this http://en.wikipedia.org/wiki/Catalan_number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359038723356:dwBut how did u get

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2reduce an to a(n1) ... then a(n1) to a(n2) .. until a1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0an=(2n3)/2n a(n1) an=(2n3)/2n *(2n5)/(2n2) * (2n7)/(2n4) *...*3/6 * 1/4 *1/2

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2(2n3)(2n5) ... 5.3.1 ^ there are holes ... fill it (2n2)(2n4) ... put the even terms at the top and at the bottom. make 2n! at the top and collect 2 from these at the denominator.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did a algebra mistake earlier

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanx for your help @experimentX
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