Question Number 104701 by abony1303 last updated on 23/Jul/20

Commented byabony1303 last updated on 23/Jul/20

Pls help

Answered by mathmax by abdo last updated on 23/Jul/20

A_n =(1/n)ln((((2n)!)/(n! n^n )))  we have n! ∼n^n  e^(−n) (√(2πn))(n →+∞)(stirling) ⇒  (2n)! ∼(2n)^(2n)  e^(−2n) (√(4πn)) ⇒(((2n)!)/(n! n^n )) ∼(((2n)^(2n)  e^(−2n) .2(√(πn)))/(n^n  e^(−n) (√(2πn))n^n ))  =(√2)×2^(2n)  e^(−n)  ⇒ A_n =(1/n)ln((√2)×2^(2n) ×e^(−n) )=((ln2)/(2n)) +((2n)/n)ln(2)−(n/n)  =((ln2)/(2n)) +2ln2−1 ⇒lim_(n→+∞)  A_n =2ln2 −1

Commented byabony1303 last updated on 23/Jul/20

couldnt catch you ser :(

Commented bymathmax by abdo last updated on 23/Jul/20

what do you mean with this...

Commented byabony1303 last updated on 23/Jul/20

didn′t understand at some points from  beginning. can u pls write some  descriptions?

Commented byabony1303 last updated on 24/Jul/20

Thank you ser. It was very helpful.  Didnt′t know about Sterling′s  approximation.

Commented bymathmax by abdo last updated on 24/Jul/20

here i have used th stirling formula  n! ∼ n^n  e^(−n) (√(2πn))  (for n →+∞)   i think the answer is clear