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sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
dw:1358686236762:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
The R.H.S=1/2
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
It is dw:1358696599099:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
Not (2n2)*(n1)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
It is (2n2)C(n1)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
oh, \[^{2n2}C_{n1}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
dw:1358757323764:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
dw:1358757562194:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
I think this is untrue, the LHS diverges while RHS converges
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
How can we say LHS diverges?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
the LHS is of order n, definitely it diverges
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
RHS is convergent as per p test
 one year ago

experimentX Group TitleBest 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.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
The actual question is here http://openstudy.com/users/jonask#/updates/50ed3bece4b0d4a537cd13df
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
LHS is divergent by comparison test
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
So, it is false?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
dw:1358758646393:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
let me check this sequence with mathematica
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
what result did u get from mathematica?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
So fat the result is positive for that sequence.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
up to how many terms u tried?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
for 1000 the result is 0.982152
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.2
Here'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}]
 one year ago

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

youarestupid Group TitleBest ResponseYou've already chosen the best response.0
could somebody give a medal please
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
medal to whom?
 one year ago

youarestupid Group TitleBest ResponseYou've already chosen the best response.0
I gave you one
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
If u will prove it may get more than 1 medal
 one year ago

youarestupid Group TitleBest ResponseYou've already chosen the best response.0
I am only I 6 grade I havent done anything like that before I am sorry. but could you please give me a medal
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
you 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
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
dw:1359038723356:dwBut how did u get
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
reduce an to a(n1) ... then a(n1) to a(n2) .. until a1
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
an=(2n3)/2n a(n1) an=(2n3)/2n *(2n5)/(2n2) * (2n7)/(2n4) *...*3/6 * 1/4 *1/2
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
then?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
collect 2 first
 one year ago

experimentX Group TitleBest 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.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
Oh wait...I got it
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
I did a algebra mistake earlier
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.4
Thanx for your help @experimentX
 one year ago
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