Question

Calculus 2: Sum from n=1 to infinity of (n^4 + ((-1)^n)*n^2) / ((-n)^5 + 3(n^3))

Answer 1

Answer:

(n^4 + ((-1)^n)*n^2) / ((-n)^5 + 3(n^3))

  n^4 + -1/ -1^5 + 3n^3

  n^4 + -1^n / -1^5 +3^3

  n^4/ -1  + 3n^3

  n^4/ -1 + 3x1^3

  n/-1 + 3n

  1/-1+3

  n = 2