Evaluate the product n = 3 ∏ ∞ n 6 − 6 4 ( n 3 + 3 n ) 2 .
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Define the sequence { x n } as, x n = n 6 − 6 4 ( n 3 + 3 n ) 2 The idea is to decompose x n into simple fractions that might telescope. So we rewrite the denominator of the general term, as n 6 − 6 4 = ( n 3 + 8 ) ( n 3 − 8 ) = ( n + 2 ) ( n 2 − 2 n + 4 ) ( n − 2 ) ( n 2 + 2 n + 4 ) = ( n + 2 ) ⋅ ( ( n − 1 ) 2 + 3 ) ⋅ ( n − 2 ) ⋅ ( ( n + 1 ) 2 + 3 ) Since the denominator is a product of four terms, we shall try doing the same for the numerator. ( n 3 + 3 n ) 2 = ( n ( n 2 + 3 ) ) 2 = n 2 ( n 2 + 3 ) 2 = n ⋅ n ⋅ ( n 2 + 3 ) ⋅ ( n 2 + 3 ) So, the general term of the sequence is, x n = n 6 − 6 4 ( n 3 + 3 n ) 2 = ( n + 2 ) ⋅ ( ( n − 1 ) 2 + 3 ) ⋅ ( n − 2 ) ⋅ ( ( n + 1 ) 2 + 3 ) n ⋅ n ⋅ ( n 2 + 3 ) ⋅ ( n 2 + 3 ) = ( n + 2 n ) ⋅ ( n − 2 n ) ⋅ ( ( n + 1 ) 2 + 3 n 2 + 3 ) ⋅ ( ( n − 1 ) 2 + 3 n 2 + 3 ) We shall use the following infinite products: n = 3 ∏ ∞ n + 2 n = k → ∞ lim n = 3 ∏ k n + 2 n = k → ∞ lim ( 5 3 × 6 4 × 7 5 × 8 6 × ⋯ × k + 1 k − 1 × k + 2 k ) = k → ∞ lim ( k + 1 ) ( k + 2 ) 1 2 n = 3 ∏ ∞ n − 2 n = k → ∞ lim n = 3 ∏ k n − 2 n = k → ∞ lim ( 1 3 × 2 4 × 3 5 × 4 6 × ⋯ × k − 3 k − 1 × k − 2 k ) = k → ∞ lim 2 ( k − 1 ) k n = 3 ∏ ∞ ( n + 1 ) 2 + 3 n 2 + 3 = k → ∞ lim n = 3 ∏ k ( n + 1 ) 2 + 3 n 2 + 3 = k → ∞ lim ( 1 9 1 2 × 2 8 1 9 × 3 9 2 8 × ⋯ × k 2 + 3 ( k − 1 ) 2 + 3 × ( k + 1 ) 2 + 3 k 2 + 3 ) = k → ∞ lim ( k + 1 ) 2 + 3 1 2 n = 3 ∏ ∞ ( n − 1 ) 2 + 3 n 2 + 3 = k → ∞ lim n = 3 ∏ k ( n − 1 ) 2 + 3 n 2 + 3 = k → ∞ lim ( 7 1 2 × 1 2 1 9 × 1 9 2 8 × ⋯ × ( k − 2 ) 2 + 3 ( k − 1 ) 2 + 3 × ( k − 1 ) 2 + 3 k 2 + 3 ) = k → ∞ lim 7 k 2 + 3 Therefore, the product of all terms of the sequence { x n } is, n = 3 ∏ ∞ x n = ( n = 3 ∏ ∞ n + 2 n ) ⋅ ( n = 3 ∏ ∞ n − 2 n ) ⋅ ( n = 3 ∏ ∞ ( n + 1 ) 2 + 3 n 2 + 3 ) ⋅ ( n = 3 ∏ ∞ ( n − 1 ) 2 + 3 n 2 + 3 ) = k → ∞ lim ( ( k + 1 ) ( k + 2 ) 1 2 ⋅ 2 ( k − 1 ) k ⋅ ( k + 1 ) 2 + 3 1 2 ⋅ 7 k 2 + 3 ) = k → ∞ lim 7 ( k 2 + 3 k + 2 ) ( k 2 + 2 k + 4 ) 7 2 ( k 2 − 1 ) ( k 2 + 3 ) = 7 7 2 ⋅ k → ∞ lim ( 1 + k 3 + k 2 2 ) ( 1 + k 2 + k 2 4 ) ( 1 − k 2 1 ) ( 1 + k 2 3 ) = 7 7 2
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n = 3 ∏ ∞ n 6 − 6 4 ( n 3 + 3 n ) 2 = n = 3 ∏ ∞ ( n − 2 ) ( n + 2 ) ( n 2 − 2 n + 4 ) ( n 2 + 2 n + 4 ) n ⋅ n ( n 2 + 3 ) ( n 2 + 3 ) = n = 3 ∏ ∞ ( n − 2 ) ( n + 2 ) ( ( n − 1 ) 2 + 3 ) ( ( n + 1 ) 2 + 3 ) n ⋅ n ( n 2 + 3 ) ( n 2 + 3 ) = n = 3 ∏ ∞ n − 2 n ⋅ n = 3 ∏ ∞ n + 2 n ⋅ n = 3 ∏ ∞ ( n − 1 ) 2 + 3 n 2 + 3 ⋅ n = 3 ∏ ∞ ( n + 1 ) 2 + 3 n 2 + 3 = n = 1 ∏ ∞ n n + 2 ⋅ n = 3 ∏ ∞ n + 2 n ⋅ n = 2 ∏ ∞ n 2 + 3 ( n + 1 ) 2 + 3 ⋅ n = 3 ∏ ∞ ( n + 1 ) 2 + 3 n 2 + 3 = 1 ⋅ 2 3 ⋅ 4 n = 3 ∏ ∞ ( n n + 2 ⋅ n + 2 n 1 ) ⋅ 7 1 2 n = 3 ∏ ∞ ( n 2 + 3 ( n + 1 ) 2 + 3 ⋅ ( n + 1 ) 2 + 3 n 2 + 3 1 ) = 7 7 2