n → ∞ lim k = 1 ∑ n n 2 n 2 − k 2 = ?
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Nice solution. A shortcut would be to note that ∫ 0 1 1 − x 2 d x is the area of a quarter of a unit circle.
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Precisely what I did. :-)
Right... Thanks
Good shortcut
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n → ∞ lim n 1 k = 1 ∑ n 1 − ( n k ) 2
Using Reimann Sums , this is:
∫ 0 1 1 − x 2 d x
Substitute x = sin θ so that integration is:
∫ 0 2 π cos θ × cos θ d θ
= ∫ 0 2 π 2 1 + cos 2 θ d θ
= [ 2 θ + 2 sin 2 θ ] 0 2 π = 4 π