Converting a function to a power series with only variables. A bit stuck

Question

I am trying to convert this function into a power series and I'm a tad stuck on where to go from here:

$$f(x) = \frac{x+a}{x^2 + a^2}$$ and a $\gt 0$

Are these steps valid:

$$(x+a) \cdot \frac{1}{a^2 + x^2}$$

$$ = \frac{(x+a)}{a^2} \cdot \frac{1}{1 + (\frac{x}{a})^2}$$

$$ = \frac{(x+a)}{a^2} \cdot \frac{1}{1 - (\frac{-x}{a})^2}$$

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

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

But I'm stuck on how to incorporate the fraction on the outside back in.

Answer 1

Continuing from before a mistake was made, $$\begin{align} \frac{x+a}{a^2} \cdot \frac{1}{1 + \frac{x^2}{a^2}} &=\frac{x+a}{a^2} \cdot \frac{1}{1 -\left(-\frac{x^2}{a^2}\right)}\\ &=\frac{x+a}{a^2} \sum_{k=0}^\infty \left(-\frac{x^2}{a^2}\right)^k\\ &=\frac{x+a}{a^2} \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{a^{2k}}\\ &= \sum_{k=0}^\infty \frac{(x+a)\left((-1)^kx^{2k}\right)}{a^{2k+2}}\\ &= \sum_{k=0}^\infty \left(\frac{(-1)^kx^{2k+1}}{a^{2k+2}}+\frac{(-1)^kx^{2k}}{a^{2k+1}}\right)\\ &= \sum_{k=0}^\infty \frac{(-1)^{\left\lfloor\frac{k}2\right\rfloor}}{a^{k+1}}x^k\\ \end{align}$$

Answer 2

Note that\begin{align}\frac1a-\frac{x^2}{a^3}+\frac{x^4}{a^5}-\frac{x^6}{a^7}+\cdots&=\frac1a\left(1-\left(\frac xa\right)^2+\left(\frac xa\right)^4-\left(\frac xa\right)^6+\cdots\right)\\&=\frac1a\cdot\frac1{1+\frac{x^2}{a^2}}\\&=\frac a{x^2+a^2}\end{align}if $\left\lvert\frac xa\right\rvert<1$. A similar computation shows that$$\frac x{a^2}-\frac{x^3}{a^4}+\frac{x^5}{a^6}-\frac{x^7}{a^8}+\cdots=\frac x{x^2+a^2},$$again, if $\left\lvert\frac xa\right\rvert<1$. Therefore,$$\left\lvert\frac xa\right\rvert<1\implies\frac{x+a}{x^2+a^2}=\frac1a+\frac x{a^2}-\frac x{a^3}-\frac{x^3}{a^4}+\cdots$$

Answer 3

Rewrite the fraction as $$f(x) = \frac{x+a}{x^2 + a^2}=\frac 1a \frac{1+\cfrac xa}{1+\cfrac{x^2}{a^2}}$$ and set $t=\dfrac xa$. You have to find the expansion of \begin{align} \frac{1+t}{1+t^2}&=(1+t)\,\frac 1{1+t^2}\\ &=(1+t)(1-t^2+t^4-t^6+t^8-\dotsm)\\ &=1-t^2+t^4-t^6+t^8+\dots\\ &\quad\, +t-t^3+t^5-t^7+t^9-\dotsm \\ &=(1+t)-(t^2+t^3)+(t^4+t^5)-(t^6+t^7)+\dotsm \end{align} Can you find a general formula from these elements?