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Rachel Villa

Rachel Villa

Answered question

2022-05-21

Calculate the sum
n = 1 ( 2 n 1 ) ! ! ( 2 n ) ! ! 2 n
where ( 2 n 1 ) ! ! = 1 3 ( 2 n 1 ), ( 2 n ) ! ! = 2 4 2 n

Answer & Explanation

rideonthebussp

rideonthebussp

Beginner2022-05-22Added 10 answers

First note that
( 2 n 1 ) ! ! = ( 2 n ) ! 2 n n !
and ( 2 n ) ! ! = 2 n n !
( 2 n 1 ) ! ! ( 2 n ) ! ! 2 n = ( 2 n ) ! 2 n n ! 2 n n ! 2 n = 1 2 3 n ( 2 n n ) .
Now use the power series
(1) 1 1 4 x = n 0 ( 2 n n ) x n ,
the generating function for the central binomial coefficients. ( 2 n n ) 4 n for n 1, so (1) certainly converges at x = 1 8 , and we have
2 = 1 1 1 / 2 = n 0 ( 2 n n ) ( 1 8 ) n = n 0 1 2 3 n ( 2 n n ) = 1 + n 1 ( 2 n 1 ) ! ! ( 2 n ) ! ! 2 n .
wanaopatays

wanaopatays

Beginner2022-05-23Added 5 answers

Since ( 2 n ) ! ! = 2 n n ! and ( 2 n 1 ) ! ! = ( 2 n 1 ) ( 2 n 3 ) 5 3 1
n = 1 ( 2 n 1 ) ! ! ( 2 n ) ! ! 2 n = n = 1 ( 2 n 1 ) ( 2 n 3 ) 5 3 1 2 n n ! 1 2 n = n = 1 ( n 1 2 ) ( n 3 2 ) ( 3 2 ) ( 1 2 ) n ! 1 2 n = n = 1 ( 1 2 n + 1 ) ( 1 2 n + 2 ) ( 1 2 ) n ! ( 1 2 ) n = n = 0 ( 1 / 2 n ) ( 1 2 ) n 1 = ( 1 1 / 2 ) 1 / 2 1 = 2 1.

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