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The infinite sum sum_{n=1}^{infinity} 1/(n(n+2)(n+4)) equals m/n in…
Question
The infinite sum sum_{n=1}^{infinity} 1/(n(n+2)(n+4)) equals m/n in lowest terms, where m and n are coprime positive integers. Find m + n.
✓ Verified answer: 107checked by our engine — not a guess
Step-by-step solution
Partial fractions: 1/(n(n+2)(n+4)) = (1/8)/n - (1/4)/(n+2) + (1/8)/(n+4).
Group as (1/8)[1/n - 1/(n+2)] - (1/8)[1/(n+2) - 1/(n+4)]; both telescope.
sum [1/n - 1/(n+2)] = 1 + 1/2 = 3/2 and sum [1/(n+2) - 1/(n+4)] = 1/3 + 1/4 = 7/12.
So total = (1/8)(3/2) - (1/8)(7/12) = (1/8)(18/12 - 7/12) = (1/8)(11/12) = 11/96.
m/n = 11/96, so m + n = 107.
Final answer107
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