Power series for $f(x) = \frac{4}{x+2}$

Question

Find the power series $f(x) = 4/(x+2)$

We know the geometric series:

$$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

$(x+2) = 1 - (-x - 1)$

So:

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

Multiply by $4$

$$\sum_{n=1}^{\infty} 4(-1)^{n-1} \cdot (x+1)^{n+1} = \frac{4}{x+2}$$

But this isnt the correct answer as the book points out. why?

Answer 1

HINT

$$ \frac{4}{2+x}=\frac{2}{1+x/2}=\sum_{n=0}^{\infty}(-1)^n x^n 2^{1-n}\qquad\text{for}\; |x|<2 $$

Answer 2

For $\;|x|<2\;$ we get

$$\frac4{x+2}=\frac2{1+\frac x2}=2\left(1-\frac x2+\frac{x^2}{2^2}-\ldots\right)=2\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^n}=\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^{n-1}}$$

Answer 3

Given the power series you have first written:

$$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

If we rewrite the fraction:

$$\frac{4}{2+x} = \frac{2}{1+\frac{x}{2}}$$

Now we have in a form that is similar to the first power series above, $\frac{1}{1-x}$

Notice that:

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

Answer 4

Hint:-

$\dfrac{4}{x+2}=4(x+2)^{-1}$

Now expand by Binomial Theorem keeping in mind the restrictions on $x$.