Here's the question you clicked on:
sauravshakya
Can someone PROVE this:
|dw:1358686236762:dw|
It is |dw:1358696599099:dw|
It is (2n-2)C(n-1)
oh, \[^{2n-2}C_{n-1}\]
|dw:1358757323764:dw|
|dw:1358757562194:dw|
I think this is untrue, the LHS diverges while RHS converges
How can we say LHS diverges?
the LHS is of order n, definitely it diverges
RHS is convergent as per p test
\[ \frac{2 (n-1)^2}{8n} = \frac{n^2 - 2n + 1}{4n} = \left( \frac{n}{4} - \frac{1}{2} \right) + \frac{1}{4n}\] Both part of above diverges.
The actual question is here http://openstudy.com/users/jonask#/updates/50ed3bece4b0d4a537cd13df
LHS is divergent by comparison test
|dw:1358758646393:dw|
let me check this sequence with mathematica
what result did u get from mathematica?
So fat the result is positive for that sequence.
up to how many terms u tried?
for 1000 the result is 0.982152
The compact form of the recurrence relation is \[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)}\]
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}]
\[ a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)} = \frac{1}{2^{2n}(2n-1)} \times \binom{2n}{n} \]
could somebody give a medal please
If u will prove it may get more than 1 medal
I am only I 6 grade I havent done anything like that before I am sorry. but could you please give me a medal
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
|dw:1359038723356:dw|But how did u get
reduce an to a(n-1) ... then a(n-1) to a(n-2) .. until a1
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
(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.
Oh wait...I got it
I did a algebra mistake earlier
Thanx for your help @experimentX