limits

Stirling^s formula

i) log( r_(n 1) / r_n ) = -1 (n ½) log(1 1/n)

ii) Applying l^Hopital Rule twice, we have
lim_(x→0) (((1 x/2) log(1 x) -x ) / (x^3) ) = 1/12

iii) Let a_n = log( r_(n 1) / r_n ) , then from (i), we have
a_n
= -1 (n ½) log(1 1/n)
= n ( -1/n ( 1 (1/n)/2 ) log(1 1/n) )
Hence,
a_n / (1/n^2) = ( -1/n ( 1 (1/n)/2 ) log(1 1/n) ) / (1/n^3)
and then
lim_(n→∞) a_n / (1/n^2) = 1/12 ... by (ii) ....(*)
Since Σ(1/n^2) converges absolutely, (*) implies that Σa_n converges absolutely by the comparison test.
Notice that
Σa_n
= Σlog( r_(n 1) / r_n ) , n=1 to N
= log( r_(N 1) / r_1 )
This just means that the limit lim_(n→∞) log( r_(n 1) / r_1 ) exists
and hence
lim_(n→∞) exp( log( r_(n 1) / r_1 ) )
= lim_(n→∞) ( r_(n 1) / r_1 ) )
exists since exp( ) is a continuous function.
This has just shown that lim_(n→∞) r_n exists.

2008-03-09 12:12:34 補充：
Stirling^s formula

i) log( r_(n 1) / r_n ) = -1 plus (n plus ½) log(1 plus 1/n)

ii) Applying l^Hopital Rule twice, we have
lim_(x→0) (((1 plus x/2) log(1 plus x) -x ) / (x^3) ) = 1/12

2008-03-09 12:14:50 補充：
iii) Let a_n = log( r_(n 1) / r_n ) , then from (i), we have
a_n
= -1 plus (n plus ½) log(1 plus 1/n)
= n times ( -1/n plus ( 1 plus (1/n)/2 ) log(1 plus 1/n) )
Hence,
a_n / (1/n^2) = ( -1/n plus ( 1 plus (1/n)/2 ) log(1 plus 1/n) ) / (1/n^3)
and then
lim_(n→∞) a_n / (1/n^2) = 1/12 ... by (ii) ....(*)

2008-03-09 12:17:17 補充：
Notice that
Σa_n
= Σlog( r_(n plus 1) / r_n ) , n=1 to N
= log( r_(N plus 1) / r_1 )
This just means that the limit lim_(n→∞) log( r_(n plus 1) / r_1 ) exists
and hence
lim_(n→∞) exp( log( r_(n plus 1) / r_1 ) )
= lim_(n→∞) ( r_(n plus 1) / r_1 ) )