What is the sum of ∑1/(n^3 (n + 1)^3) for n = 1 to ∞ and why?

Σ(n = 1 to ∞) 1/(n³ (n+1)³)

= Σ(n = 1 to ∞) [6/n - 3/n² + 1/n³ - 6/(n+1) - 3/(n+1)² - 1/(n+1)³]

by partial fractions

= Σ(n = 1 to ∞) [6/n - 6/(n+1)] + Σ(n = 1 to ∞) [1/n³ - 1/(n+1)³]

- 3 Σ(n = 1 to ∞) 1/n² - 3 Σ(n = 1 to ∞) 1/(n+1)²

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The first two terms are telescoping series

Σ(n = 1 to ∞) [6/n - 6/(n+1)]

= lim(k→∞) (6/1 - 6/(k+1))

= 6.

Σ(n = 1 to ∞) [1/n³ - 1/(n+1)³]

= lim(k→∞) (1/n³ - 1/(n+1)³)

= 1.

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As for the last two terms:

-3 Σ(n = 1 to ∞) 1/n² - 3 Σ(n = 1 to ∞) 1/(n+1)²

= -3 Σ(n = 1 to ∞) 1/n² - 3 Σ(n = 2 to ∞) 1/n², index shift

= -3 Σ(n = 1 to ∞) 1/n² - 3 [-1 + Σ(n = 1 to ∞) 1/n²]

= 3 - 6 Σ(n = 1 to ∞) 1/n²

= 3 - 6 * (π²/6), since Σ(n = 1 to ∞) 1/n² = π²/6

= 3 - π².

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Therefore, the series in question equals

6 + 1 + (3 - π²) = 10 - π².

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Fun problem!

Do partial fractions, which can be very tedious, use RiemannZeta(2) = π^2 /6. You should get (10 - π^2) as the answer.

i guess u must make numerator as (n+1)-n then split the fraction then it becomes doable..not sure..however go to link...may be some pretty hard integrals..

bcoz pi^2 in answer