Problem #1: Infinity and infinity combined

Calculus Level 4

f ( k ) = ∑ n = 1 ∞ 1 n ( n + 1 ) ( n + 2 ) ⋯ ( n + k ) , f(k)=\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)\cdots(n+k)},

For positive integer k k , f ( k ) f(k) is defined as above. Then ∑ n = 5 ∞ n ( n − 1 ) f ( n ) = p q \displaystyle \sum_{n=5}^{\infty} n(n-1)f(n) = \frac{p}{q} , where p p and q q are coprime positive integers. Find the value of p 2 + q 2 p^2+q^2 .

This problem is a part of the series <Advanced Problems> .


The answer is 577.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Oct 17, 2018

Since

1 n ( n + 1 ) ( n + 2 ) ⋯ ( n + k ) = 1 k ( 1 n ( n + 1 ) ( n + 2 ) ⋯ ( n + k − 1 ) − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ⋯ ( n + k ) ) \frac{1}{n(n+1)(n+2)\cdots(n+k)}=\frac{1}{k}\left(\frac{1}{n(n+1)(n+2)\cdots(n+k-1)}-\frac{1}{(n+1)(n+2)(n+3)\cdots(n+k)}\right)

we may have

f ( k ) = ∑ n = 1 ∞ 1 k ( 1 n ( n + 1 ) ( n + 2 ) ⋯ ( n + k − 1 ) − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ⋯ ( n + k ) ) = 1 k ⋅ k ! f(k) = \sum_{n=1}^{\infty} \frac{1}{k}\left(\frac{1}{n(n+1)(n+2)\cdots(n+k-1)}-\frac{1}{(n+1)(n+2)(n+3)\cdots(n+k)}\right) = \frac{1}{k\cdot k!}

Hence

∑ n = 5 ∞ n ( n − 1 ) f ( n ) = ∑ n = 5 ∞ n − 1 n ! = ∑ n = 5 ∞ ( 1 ( n − 1 ) ! − 1 n ! ) = 1 24 . \sum_{n=5}^{\infty} n(n-1)f(n) \\ = \sum_{n=5}^{\infty} \frac{n-1}{n!} \\ =\sum_{n=5}^{\infty}\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right) \\ =\frac{1}{24}.

Therefore, p 2 + q 2 = 1 2 + 2 4 2 = 577 . p^2+q^2 = 1^2+24^2 = \boxed{577}.

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