i → ∞ lim ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ j = 1 ∑ j = i ( ( j + 1 ) ! ) 2 ⎝ ⎜ ⎜ ⎛ ⎝ ⎜ ⎜ ⎛ ∑ k = 1 k = j lim l → ∞ ∑ m = 1 m = l ∏ n = 0 n = k m + n 1 1 ⎠ ⎟ ⎟ ⎞ + 1 ⎠ ⎟ ⎟ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
If the answer can be expressed as e a − b , select a + b .
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You could've done the first part like this.
l → ∞ lim m = 1 ∑ m = l ∏ n = 0 n = k m + n 1 l → ∞ lim m = 1 ∑ m = l m × ( m + 1 ) × ( m + 2 ) … ( m + k ) 1 l → ∞ lim m = 1 ∑ m = l k 1 m × ( m + 1 ) × ( m + 2 ) … ( m + k ) ( m + k ) − m l → ∞ lim k 1 m = 1 ∑ m = l m × ( m + 1 ) … ( m + k − 1 ) 1 − ( m + 1 ) × … ( m + k ) 1 l → ∞ lim k 1 ( 1 × ( 2 ) × ( 3 ) … ( k ) 1 − ( l + 1 ) × ( l + 2 ) … ( l + k ) 1 ) k ( k ! ) 1
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Oh yeah, it's just the same thing that Kishlaya's identity has 2 proofs - one relating to Beta function and the other(and more clever) using telescoping.
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Nice easy problem!
Well, we'll start with the product at the bottom.
∏ n = 0 k ( m + n ) 1 = ( m + k ) ! ( m − 1 ) ! which is quite easy to observe.
Now the next summation,
l → ∞ lim m = 1 ∑ l Γ ( m + k + 1 ) Γ ( m )
Multiply and divide by Γ ( k + 1 ) and take out the denominator while let the numerator be inside the summation.
Γ ( k + 1 ) 1 l → ∞ lim m = 1 ∑ l Γ ( m + k + 1 ) Γ ( m ) Γ ( k + 1 )
Then we can substitute Beta function inside summation,
Γ ( k + 1 ) 1 l → ∞ lim m = 1 ∑ l ∫ 0 1 t m − 1 ( 1 − t ) k
∫ 0 1 ( 1 − t ) k m = 1 ∑ ∞ t m − 1
∫ 0 1 ( 1 − t ) k − 1 = k 1
Putting everything in the second summation,
k = 1 ∑ j k . k ! 1 1
k = 1 ∑ j k . k ! = k = 1 ∑ j ( k + 1 ) ! − k !
= ( j + 1 ) ! − 1
Finally into the last summation,
j = 1 ∑ ∞ ( j + 1 ) ! 1
e − 2