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# Help with infinite series?

∑ ((3 / (k-1)^2) - (3 / k^2))
k=2

Oh that was simple...I was over thinking it by trying to find the limit

3 years ago

by sahsjing
Member since:
20 December 2006
Total points:
230,899 (Level 7)

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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• by hayharbr
Member since:
13 February 2006
Total points:
327,202 (Level 7)
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
• 2 people rated this as good
• Member since:
12 May 2009
Total points:
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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