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caculate limit
lim n-infinite Σn i=1 [8-(3i/n)^2]3/n
lim n-infinite Σn i=1 [8-(3i/n)^2]3/n
Answers
Answered by
oobleck
Σ[8-(3i/n)^2]3/n
= Σ24/n - 27i/n^3
= Σ24/n 1 - 27/n^3 Σi
Now, sum(i=1..n) i^2 = n(n+1)(2n+1)/6 so
= 24 - 27/n^3 * n(n+1)(2n+1)/6
= 24 - 9/2 (2 + 3/n + 1/n^2)
Now as n→∞ that becomes
24 - 9 = 15
Note that this is just
∫[0..3] 8-x^2 dx = 8x - 1/3 x^3 [0..3] = 15
= Σ24/n - 27i/n^3
= Σ24/n 1 - 27/n^3 Σi
Now, sum(i=1..n) i^2 = n(n+1)(2n+1)/6 so
= 24 - 27/n^3 * n(n+1)(2n+1)/6
= 24 - 9/2 (2 + 3/n + 1/n^2)
Now as n→∞ that becomes
24 - 9 = 15
Note that this is just
∫[0..3] 8-x^2 dx = 8x - 1/3 x^3 [0..3] = 15