Partial Fractions May Help Here!

Calculus Level 4

The expression ∑ n = a ∞ ∑ b = n ∞ 1 ( n − a ) ! ( b − n ) ! \displaystyle \sum_{n=a}^{\infty} \sum_{b=n}^{\infty} \frac{1}{(n-a)!(b-n)!} , for a , b , n ∈ N a,b,n \in N , can be expressed as k 2 k^2 . Find k k .

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The answer is 2.71828.

Hasan Kassim by Hasan Kassim , 22, Lebanon

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1 solution

Assuming that the question meant ∑ n = a ∞ ∑ b = n ∞ 1 ( n − a ) ! ( b − n ) ! \sum_{n=a}^\infty \sum_{b=n}^\infty\frac{1}{(n-a)!(b-n)!} We can solve this problem. First start by pulling out the 1 / ( n − a ) ! 1/(n-a)! part and manipulating indexes ∑ n = a ∞ ∑ b = n ∞ 1 ( n − a ) ! ( b − n ) ! = ∑ n = a ∞ 1 ( n − a ) ! ∑ b = n ∞ 1 ( b − n ) ! = ∑ n − a = 0 ∞ 1 ( n − a ) ! ∑ b − n = 0 ∞ 1 ( b − n ) ! = ∑ n = 0 ∞ 1 n ! ∑ k = 0 ∞ 1 k ! = e ⋅ e = e 2 \sum_{n=a}^\infty \sum_{b=n}^\infty\frac{1}{(n-a)!(b-n)!}=\sum_{n=a}^\infty \frac1{(n-a)!}\sum_{b=n}^\infty\frac1{(b-n)!}\\=\sum_{n-a=0}^\infty \frac1{(n-a)!}\sum_{b-n=0}^\infty\frac1{(b-n)!}=\sum_{n=0}^\infty \frac1{n!}\sum_{k=0}^\infty\frac1{k!}=e\cdot e=e^2 Thus the answer is e ≈ 2.718 e\approx\boxed{2.718}

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I think that the index summation is also incorrect. Likely what he wanted was

∑ n = a ∞ ∑ m = b ∞ 1 ( n − a ) ! ( m − b ) ! \sum_{n=a}^\infty \sum_{m=b}^\infty\frac{1}{(n-a)!(m-b)!}

Calvin Lin Staff - 6 years, 8 months ago

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Oh, excuse me, I will edit the problem..Sorry for that mistake.

Hasan Kassim - 6 years, 8 months ago

Sorry, I did not understand how did you get 1 n ! \frac{1}{n!} from 1 ( n − a ) ! \frac{1}{(n-a)!} ?

Arpit Agarwal - 6 years, 3 months ago

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