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

Step-by-step solution

1. 1

   Start from a series you already know. The geometric series gives, for $|x|<1$,

   $\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots=\sum_{n=0}^{\infty}(-1)^nx^n$

   Rather than computing seven derivatives of $(1+x)^{-2}$ at $0$, notice how the target is related to this one.

2. 2

   Differentiate both sides once. Since $\dfrac{d}{dx}\left(\dfrac{1}{1+x}\right)=-\dfrac{1}{(1+x)^2}$, differentiating term by term and negating gives the target directly:

   $-\frac{1}{(1+x)^2}=-1+2x-3x^2+4x^3-\cdots\;\Longrightarrow\;\frac{1}{(1+x)^2}=1-2x+3x^2-4x^3+\cdots$

   Term-by-term differentiation is legitimate inside the radius of convergence, which stays $|x|<1$.

3. 3

   Write the general coefficient. The pattern is a sign that alternates and a magnitude that counts up:

   $\frac{1}{(1+x)^2}=\sum_{n=0}^{\infty}(-1)^n(n+1)x^n$

   You can confirm this against the binomial series $(1+x)^{-2}=\sum\binom{-2}{n}x^n$, since $\binom{-2}{n}=(-1)^n(n+1)$.

4. 4

   List the terms through degree 7.

   $\frac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7+O\!\left(x^8\right)$

5. 5

   Check a value and note the convergence limit. At $x=0.1$ the truncation gives $1-0.2+0.03-0.004+0.0005-0.00006+0.000007-0.0000008=0.8264462$, and the exact value is $\dfrac{1}{1.1^2}=0.826446\overline{28}$ $\checkmark$. At $x=1$ the series becomes $1-2+3-4+\cdots$, which does not converge, matching the radius $|x|<1$ inherited from the geometric series.