Let f (x) = (x -arctanx) /(x ^3 )for each x ≠ 0 Find the power series with sum equal to f (x) (when | x | ≤ 1, x ≠ 0), and use this power series to calculate the limit limx → 0f (x).
Yes i did, i got series (-1)^(n+1) *x^(2n-2) (series from n=1 to infinity)
But it was not the correct answer
The correctly answer is ∑k=0∞((−1)^k)/(2k+3)*x^(2k)
1/3
I do not understand how he get this answer. Have you any suggestion?
$\displaystyle x-\arctan x = x - \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n+1}}{2n+1}$
The first term of the $\displaystyle \arctan x$ series is x. So, when you perform the subtraction, you get:
$\displaystyle x-\arctan x = \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n+3}}{2n+3}$
Then, dividing by $\displaystyle x^3$ gives the series
$\displaystyle \dfrac{x-\arctan x}{x^3} = \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n}}{2n+3}$