anonymous
  • anonymous
Can someone PROVE this:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1358686236762:dw|
anonymous
  • anonymous
The R.H.S=1/2
anonymous
  • anonymous
It is |dw:1358696599099:dw|

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More answers

anonymous
  • anonymous
Not (2n-2)*(n-1)
anonymous
  • anonymous
It is (2n-2)C(n-1)
UnkleRhaukus
  • UnkleRhaukus
oh, \[^{2n-2}C_{n-1}\]
anonymous
  • anonymous
yep
experimentX
  • experimentX
|dw:1358757323764:dw|
anonymous
  • anonymous
|dw:1358757562194:dw|
experimentX
  • experimentX
I think this is untrue, the LHS diverges while RHS converges
anonymous
  • anonymous
How can we say LHS diverges?
experimentX
  • experimentX
the LHS is of order n, definitely it diverges
anonymous
  • anonymous
RHS is convergent as per p test
experimentX
  • experimentX
\[ \frac{2 (n-1)^2}{8n} = \frac{n^2 - 2n + 1}{4n} = \left( \frac{n}{4} - \frac{1}{2} \right) + \frac{1}{4n}\] Both part of above diverges.
anonymous
  • anonymous
anonymous
  • anonymous
LHS is divergent by comparison test
anonymous
  • anonymous
So, it is false?
anonymous
  • anonymous
ya
anonymous
  • anonymous
|dw:1358758646393:dw|
experimentX
  • experimentX
let me check this sequence with mathematica
anonymous
  • anonymous
what result did u get from mathematica?
experimentX
  • experimentX
So fat the result is positive for that sequence.
experimentX
  • experimentX
*far
anonymous
  • anonymous
up to how many terms u tried?
experimentX
  • experimentX
for 1000 the result is 0.982152
experimentX
  • experimentX
The compact form of the recurrence relation is \[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)}\]
experimentX
  • experimentX
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}]
experimentX
  • experimentX
\[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)} = \frac{1}{2^{2n}(2n-1)} \times \binom{2n}{n} \]
anonymous
  • anonymous
could somebody give a medal please
anonymous
  • anonymous
medal to whom?
anonymous
  • anonymous
me
anonymous
  • anonymous
I gave you one
anonymous
  • anonymous
If u will prove it may get more than 1 medal
anonymous
  • anonymous
I am only I 6 grade I havent done anything like that before I am sorry. but could you please give me a medal
experimentX
  • experimentX
you can show that \[ \sum_{n=1}^\infty \frac{1}{2^{2n}(2n-1)} \times \binom{2n}{n} = 1 \] using this http://en.wikipedia.org/wiki/Catalan_number
experimentX
  • experimentX
http://tinyurl.com/bfj7oep
anonymous
  • anonymous
|dw:1359038723356:dw|But how did u get
experimentX
  • experimentX
reduce an to a(n-1) ... then a(n-1) to a(n-2) .. until a1
anonymous
  • anonymous
an=(2n-3)/2n a(n-1) an=(2n-3)/2n *(2n-5)/(2n-2) * (2n-7)/(2n-4) *...*3/6 * 1/4 *1/2
anonymous
  • anonymous
then?
experimentX
  • experimentX
collect 2 first
experimentX
  • experimentX
(2n-3)(2n-5) ... 5.3.1 ^ there are holes ... fill it (2n-2)(2n-4) ... 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
  • anonymous
Oh wait...I got it
anonymous
  • anonymous
I did a algebra mistake earlier
anonymous
  • anonymous
Thanx for your help @experimentX

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