Finding a geometric series for $\frac{1}{(1 + x)^2}$

Question

I'm asked to find a geometric series for $f(x) = \frac{1}{(1 + x)^2}$, I integrate it first and I get $F(x) = \int{\frac{1}{(1 + x)^2}dx} = \frac{1}{-2(1 - (-x))}$

Since $\frac1{1-x} = \sum_{n=0}^\infty x^n$, $F(x) = \frac{-1}2\sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{2}x^n$

If I derive that, $f(x) = \sum_{n=0}^\infty\frac{(-1)^{n+1} n x^{n-1}}{2} = \sum_{n=1}^\infty\frac{(-1)^n (n + 1) x^n}{2}$

But my textbook says that $f(x) = \sum_{n=0}^\infty (-1)^n (n + 1) x^n$

Obviously those 2 things are pretty different (notice the index). What did I get wrong?

Answer 1

Original question:

find a geometric series for $$f(x) = \frac{1}{1+x^2}$$

As you wrote, $$\frac{1}{1-y} = \displaystyle\sum_{n=0}^\infty y^n$$

Substituting $y=-x^2$, we have

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

Edited question:

find a geometric series for $$g(x) = \frac{1}{\left(1+x\right)^2}$$

Recall that $$(1+x)^2 = 1+2x+x^2$$

As you wrote, $$\frac{1}{1-y} = \displaystyle\sum_{n=0}^\infty y^n$$

Substituting $y=-2x-x^2$, we have

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

Answer 2

HINT

The geometric series is given by \begin{align*} \frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \ldots = \sum_{n=0}^{\infty}x^{n} \end{align*}

Answer 3

In the step of integration, it should be $$F(x)= \int \frac{1}{(1+x)^2} \mathrm{d} x =- \frac {1}{1+x} + \mathrm{C}$$ You've got an extra factor of $\frac 12$. For setting the index correctly, you need to find $\mathrm C$ by initial value of $n$.