The value of lim_n 0.05em _k=1^n 0.05em n^3(n^2+k^2)(n^2+3k^2) is:

Question

Accepted Answer

Correct answer: B. (b) lim_n 0.05em _k=1^n 0.05em n^3n^4(1+ k^2n^2)(1+ 3k^2n^2) = lim_n 0.05em 1n _k=1^n 0.05em n^3(1+ k^2n^2)(1+ 3k^2n^2) = _0^1 0.05em dx3(1+x^2)( 13+x^2) = _0^1 0.05em 13 32 (x^2+1)-(x^2+ 13)(1+x^2)(x^2+ 13) dx = 12 _0^1 0.05em( 1x^2+ ( 1 3 )^2- 11+x^2) dx = 12 ( 3 tan^-1 ⁡ ( 3x ) )_0^1- 12 ( tan^-1 ⁡x )_0^1 = 32 ( 3 )- 12 ( 4 )= 2 3- 8 = ⁡13 8 (4 3+3 )

$The value of lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n ^{3}}{( n ^{2} + k ^{2} ) ( n ^{2} + 3 k ^{2} )} is:$

Solution

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