A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Can someone PROVE this:
anonymous
 4 years ago
Can someone PROVE this:

This Question is Closed

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

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

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

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

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

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

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

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

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

experimentX
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0The actual question is here http://openstudy.com/users/jonask#/updates/50ed3bece4b0d4a537cd13df

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

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

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

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

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

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

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

experimentX
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0could somebody give a medal please

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1359038723356:dwBut how did u get

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I did a algebra mistake earlier

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanx for your help @experimentX
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.