Find the sum of an infinite geometric series whose first term is `sum_(xto0)^(2011) ({(x)/(tanx)+2k})/(2011)` and whose common ratio is the value of `lim_(x to 0) (e^(tan^(3_(x)))-e^(x^(3)))/(2"ln"(1+x^(3)sin^(2)x))`.
[Note : where {y} denotes fractional part of y.]

Explanation

`underset(xto0)limunderset(k=1)overset(2011)sum({(x)/(tanx)+2k})/(2011)=underset(xto0)lim{(x)/(tanx)}`
`underset(xto0)lim{(x)/(tanx)}=underset(xto0)lim((x)/(tanx)-[(x)/(tanx)])` `underset(xto0)lim((x)/(tanx))=1" ":."first term is 1"`
`underset(xto0)lim(e^(x^(2))(e^(tan^(3)x-x^(3))-1))/((2ln(1+x^(3)sin^(2)x))/(x^(3)sin^(2)x)x^(3)sin^(2)x)`
`underset(xto0)lim(e^(x^(3)))/(2)(e^(tan^(3)x-^(3))-1)/(tan^(3)x-x^(3))((tan^(3)x-3))/((x^(3)sin^(2)x)/(x^(2))x^(2))`
`underset(xto0)lim((tanx-x))/ (x^(3))((tan^(2)x+x^(2)+xtanx))/(x^(2))" "(1)/(2)*(1)/(3)(3)=(1)/(2)`
Therefore common ratio is `(1)/(2)`
`S_(oo)=(1)/(1-1//2)=2`