Given that x_0 is the unique positive solution of (2−x)^(n+1)=x(x+1)⋯(x+n), try to find the asymptotic value of M=prod^n_(k=0)((k+2)/(k+x_0))^(k+2)

Josiah Owens 2022-10-20 Answered
Given that x 0 is the unique positive solution of ( 2 x ) n + 1 = x ( x + 1 ) ( x + n ), try to find the asymptotic value of
M = k = 0 n ( k + 2 k + x 0 ) k + 2
with absolute error o ( 1 ) as n , where H n denotes n-th harmonic number k = 1 n 1 / k
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Answers (1)

periasemdy
Answered 2022-10-21 Author has 15 answers
All 0-notations and o-notations work for n
First we have 2 n + 1 n ! x 0 , hence x 0 2 n + 1 / n ! = O ( 2 n / n ! ). Take logarithm, we derive that
( n + 1 ) ln ( 2 x 0 ) = ln x 0 + k = 1 n ln ( x 0 + k ) = ln n ! + ln x 0 + k = 1 n ln ( 1 + x 0 k )
therefore x 0 = 2 n + 1 ( 1 + O ( n x 0 ) ) n !
We have enough stuff to estimate ln M now.
ln M = ( n + 1 ) ln ( n + 1 ) + ( n + 2 ) ln ( n + 2 ) 2 ( n + 1 ) ln 2 + 2 n + 1 n ! 4 n n ! 2 ( n + 2 H n + 1 ) + O ( n 3 x 0 3 )
Finally, we have
exp ( 2 n + 1 n ! ) = 1 + 2 n + 1 n ! + 2 4 n n ! 2 + O ( x 0 3 ) exp ( 4 n n ! 2 ( n + 2 H n + 1 ) ) = 1 4 n n ! 2 ( n + 2 H n + 1 ) + O ( n 2 x 0 4 )
and
M = ( n + 1 ) n + 1 ( n + 2 ) n + 2 4 n + 1 ( 1 + 2 n + 1 n ! 4 n n ! 2 ( n + 2 H n 1 ) ) ( 1 + O ( n 3 x 0 3 ) )
Notice that the absolute error
O ( n 3 ( n + 1 ) n + 1 ( n + 2 ) n + 2 4 n + 1 x 0 3 )
approaches 0 when n
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