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∑ ((3 / (k-1)^2) - (3 / k^2))
k=2
Update : Oh that was simple...I was over thinking it by trying to find the limit
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This is a telescoping series.

∑ ((3 / (k-1)^2) - (3 / k^2)) = 3[1 - 1/4 + 1/4 - 1/9 + 1/9 - ...] = 3
k=2
--------
Ideas to save time: Factor out 3 first.

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  • hayharbr answered 5 years ago
    so, (3/1 - 3/4) + (3/4 - 3/9) + (3/9 - 3/16) + ...

    You can see that everything cancels except the 3/1 in the beginning and the 3 / ∞² you get in the end, which is basically 0, so the sum is just 3
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  • Slowfinger answered 5 years ago
    Sum {from 2 to ∞} (3/(k-1)² - 3/k²) =
    = 3 Sum {from k=2 to ∞} 1/(k-1)² - 3 Sum {from k=2 to ∞} 1/k²
    substitute t=k-1 in the first sum
    = 3 Sum {from t=1 to ∞} 1/t² - 3 Sum {from k=2 to ∞} 1/k²

    notice that
    Sum {from t=1 to ∞} 1/t² = π²/6
    and
    Sum {from k=2 to ∞} 1/k² = Sum {from k=1 to ∞}1/k² - 1 = π²/6 - 1

    = 3 π²/6 - 3 (π²/6 - 1)
    = 3π²/6 - 3π²/6 + 3
    = 3

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  • Help with infinite series?

    ∑ ((3 / (k-1)^2) - (3 / k^2))
    k=2
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