Find the Maclaurin series of tan(x)/(1 + x^2) up to and including the degree 7 term.

Step-by-step solution

1. 1

   Choose multiplication over repeated differentiation. Computing $f^{(7)}(0)$ directly would be brutal. Instead use two series you already know and multiply them, keeping only terms of degree $\le 7$:

   $\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+O\!\left(x^9\right)$

   $\frac{1}{1+x^2}=1-x^2+x^4-x^6+O\!\left(x^8\right)$

   The second is the geometric series $\sum(-1)^n x^{2n}$, valid for $|x|<1$.

2. 2

   Note that only odd powers can survive. $\tan x$ is odd and $\dfrac{1}{1+x^2}$ is even, so their product is odd — every even coefficient must vanish. That halves the bookkeeping and gives an instant error check.

3. 3

   Collect the $x^3$ and $x^5$ coefficients. Multiply each term of the tangent series by each term of the geometric series and gather like powers:

   $[x^3]:\; 1\cdot(-1)+\frac13\cdot 1=-\frac23$

   $[x^5]:\; 1\cdot 1+\frac13\cdot(-1)+\frac{2}{15}\cdot 1=\frac{15-5+2}{15}=\frac{12}{15}=\frac45$

4. 4

   Collect the $x^7$ coefficient. Four products contribute:

   $[x^7]:\;1\cdot(-1)+\frac13\cdot 1+\frac{2}{15}\cdot(-1)+\frac{17}{315}\cdot 1$

   Over the common denominator $315$: $\dfrac{-315+105-42+17}{315}=\dfrac{-235}{315}=-\dfrac{47}{63}$.

5. 5

   Assemble the series and test it numerically.

   $\frac{\tan x}{1+x^2}=x-\frac{2}{3}x^3+\frac{4}{5}x^5-\frac{47}{63}x^7+O\!\left(x^9\right)$

   At $x=0.1$ the truncated series gives $0.09934125873$ while $\tan(0.1)/1.01=0.09934125949$ — agreement to $8$ decimal places, exactly the accuracy an $O(x^9)$ error term predicts $\checkmark$.