第3个回答 2019-06-09
2(5) f(x) = ∑<n=1,∞>[(n^2+1)/n]x^n
= ∑<n=1,∞>nx^n + ∑<n=1,∞>x^n/n
= ∑<n=1,∞>(n+1)x^n - ∑<n=1,∞>x^n + ∑<n=1,∞>x^n/n
= [∑<n=1,∞>x^(n+1)]' - ∑<n=1,∞>x^n + ∫<0, x>[∑<n=1,∞>t^(n-1)]dt
= [x^2/(1-x)]' - x/(1-x) + ∫<0, x>dt/(1-t)
= x/(1-x)^2- x/(1-x) - ln(1-x) = x^2/(1-x)^2 - ln(1-x)
-1 < x < 1本回答被提问者采纳