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ravina
 one year ago
Best ResponseYou've already chosen the best response.0Prove that for all sufficiently large n \[ \frac{ 2n \log2 }{\log(2n+1) }  1 \ge \frac{ \log2 }{ 2 }\frac{ 2n }{ \log 2n } \]

tafkas77
 one year ago
Best ResponseYou've already chosen the best response.3Hi there! Hi there! You're going to want to post this question in the math section of OpenStudy. That way, you can be helped by people tutoring in that subject (there's a whole lot, trust me). If you post in the incorrect area, your question could be regarded as spamming, or it just may not get answered (or at least, not very quickly). Here is the link! Good luck, and enjoy OS! http://openstudy.com/study#/groups/Mathematics

hannafuller19
 one year ago
Best ResponseYou've already chosen the best response.2this is math right well you should go in the math section and ask them they might be able to help you

tafkas77
 one year ago
Best ResponseYou've already chosen the best response.3oops... didn't mean to put "Hi there!" twice :)

hannafuller19
 one year ago
Best ResponseYou've already chosen the best response.2lol i dont think they will mind.

tafkas77
 one year ago
Best ResponseYou've already chosen the best response.3If she's there... she hasn't said anything yet..

hannafuller19
 one year ago
Best ResponseYou've already chosen the best response.2ya ravina is there but he or she isnt awnsering.

tafkas77
 one year ago
Best ResponseYou've already chosen the best response.3*checks watch* Well, I better go, then. :) Thanks for helping, hanna! :)

erdog82
 one year ago
Best ResponseYou've already chosen the best response.0@ravina Please ask this in the Math section.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0shud i take it as log 10 since its just log

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0@tafkas77 did you apply for being an ambassador ? I think you will be a great amby !!

tafkas77
 one year ago
Best ResponseYou've already chosen the best response.3@AravindG wow, you do? Thanks! :) I did apply, but I heard that Preetha wanted to wait a bit before releasing the next batch. Maybe I'll apply again next time! :)
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