n → ∞ lim [ n + 1 Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n + 1 1 ) − n Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) ]
If the limit above is equal to e B A , where A and B are positive integers, find A + B .
Notation : Γ ( ⋅ ) denotes the Gamma function .
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m → 0 lim m Γ ( m ) = m → 0 lim Γ ( m + 1 ) = 1 (By gamma functional equation)
I get how you used Stolz-Cesaro the first time, but I don't get the reverse application
can you explain why this is true?
thanks
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Let L 1 be the limit in question, L 2 be the limit n → ∞ lim b n a n . And let L 3 be the limit n → ∞ lim [ ( n + 1 ) n + 1 Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) Γ ( n + 1 1 ) ÷ n n Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) ] .
By Stolz–Cesàro theorem, if L 2 exists, then so does L 1 . Also, if L 3 exists, then so does L 2 .
As b n = n and lim n → ∞ b n + 1 b n = 1 . Though the ideas (choice of b n = n ) works but as per the reverse Stolz cesaro theorem is not correct, is it?
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Let a n = n Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) and b n = n . Then the limit can be written n → ∞ lim b n + 1 − b n a n + 1 − a n .
Since b n is a strictly monotone divergent sequence, we can apply Stolz–Cesàro theorem , that is, if the following limit exists, then it must be equal to the limit in question:
n → ∞ lim b n a n = n → ∞ lim n n Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) = n → ∞ lim n n n Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) .
And again, if we apply Stolz–Cesàro theorem again, that is, if the following limit exists, then it must be equal to the limit in question:
n → ∞ lim [ ( n + 1 ) n + 1 Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) Γ ( n + 1 1 ) ÷ n n Γ ( 1 ) Γ ( 2 1 ) Γ ( 3 1 ) ⋯ Γ ( n 1 ) ] = n → ∞ lim [ Γ ( n + 1 1 ) ⋅ ( n + 1 ) n + 1 n n ] = n → ∞ lim [ ( n + 1 ) ⋅ Γ ( n + 1 1 ) ⋅ ( n + 1 n ) n ] = n → ∞ lim ⎣ ⎢ ⎢ ⎡ a n ( n + 1 ) ⋅ Γ ( n + 1 1 ) ⋅ b n ( 1 − n + 1 1 ) n + 1 ÷ c n ( 1 − n + 1 1 ) ⎦ ⎥ ⎥ ⎤
By logarithmic differentation, it's easy to show that b n → e 1 and c n → 1 . We are left to evaluate n → ∞ lim [ ( n + 1 ) ⋅ Γ ( n + 1 1 ) ] . Let m = n + 1 1 , then as n → ∞ , m → 0 , the limit is equivalent to
m → 0 lim m Γ ( m ) .
If we represent the Gamma function in terms of its Weierstrass Representation , we have
s Γ ( s ) = e − γ s k = 1 ∏ ∞ e s / k ( 1 + k s ) − 1
As s → 0 , RHS tends to 1, this can be shown by logarithmic differentation. Thus m → 0 lim m Γ ( m ) = 1 , so a n → 1 .
Putting it all together, the limit in question is equal to n → ∞ lim ( a n × b n ÷ c n ) = 1 × e 1 ÷ 1 = e 1 .
Hence, A = 1 , B = 1 ⇒ A + B = 2 .