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

tafkas77Best ResponseYou've already chosen the best response.3
Hi 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
 one year ago

hannafuller19Best ResponseYou've already chosen the best response.2
this is math right well you should go in the math section and ask them they might be able to help you
 one year ago

tafkas77Best ResponseYou've already chosen the best response.3
oops... didn't mean to put "Hi there!" twice :)
 one year ago

hannafuller19Best ResponseYou've already chosen the best response.2
lol i dont think they will mind.
 one year ago

tafkas77Best ResponseYou've already chosen the best response.3
If she's there... she hasn't said anything yet..
 one year ago

hannafuller19Best ResponseYou've already chosen the best response.2
ya ravina is there but he or she isnt awnsering.
 one year ago

tafkas77Best ResponseYou've already chosen the best response.3
*checks watch* Well, I better go, then. :) Thanks for helping, hanna! :)
 one year ago

erdog82Best ResponseYou've already chosen the best response.0
@ravina Please ask this in the Math section.
 one year ago

dan815Best ResponseYou've already chosen the best response.0
shud i take it as log 10 since its just log
 one year ago

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

tafkas77Best 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! :)
 one year ago
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