## anonymous 3 years ago Can someone PROVE this:

1. anonymous

|dw:1358686236762:dw|

2. anonymous

The R.H.S=1/2

3. anonymous

It is |dw:1358696599099:dw|

4. anonymous

Not (2n-2)*(n-1)

5. anonymous

It is (2n-2)C(n-1)

6. UnkleRhaukus

oh, $^{2n-2}C_{n-1}$

7. anonymous

yep

8. experimentX

|dw:1358757323764:dw|

9. anonymous

|dw:1358757562194:dw|

10. experimentX

I think this is untrue, the LHS diverges while RHS converges

11. anonymous

How can we say LHS diverges?

12. experimentX

the LHS is of order n, definitely it diverges

13. anonymous

RHS is convergent as per p test

14. 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.

15. anonymous

16. anonymous

LHS is divergent by comparison test

17. anonymous

So, it is false?

18. anonymous

ya

19. anonymous

|dw:1358758646393:dw|

20. experimentX

let me check this sequence with mathematica

21. anonymous

what result did u get from mathematica?

22. experimentX

So fat the result is positive for that sequence.

23. experimentX

*far

24. anonymous

up to how many terms u tried?

25. experimentX

for 1000 the result is 0.982152

26. experimentX

The compact form of the recurrence relation is $a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)}$

27. 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}]

28. experimentX

$a_n = \frac{(2n)!}{2^{2n} (n!)^2 (2n-1)} = \frac{1}{2^{2n}(2n-1)} \times \binom{2n}{n}$

29. anonymous

could somebody give a medal please

30. anonymous

medal to whom?

31. anonymous

me

32. anonymous

I gave you one

33. anonymous

If u will prove it may get more than 1 medal

34. 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

35. 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

36. experimentX
37. anonymous

|dw:1359038723356:dw|But how did u get

38. experimentX

reduce an to a(n-1) ... then a(n-1) to a(n-2) .. until a1

39. 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

40. anonymous

then?

41. experimentX

collect 2 first

42. 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.

43. anonymous

Oh wait...I got it

44. anonymous

I did a algebra mistake earlier

45. anonymous