Using the Maclaurin series for $\frac{1}{1-x}$, compute the Maclaurin series for: $$\frac{1+x}{1+x^2}$$.
What is the interval of convergence of this series?
Once you get the Maclaurin series for the function, I know how to find the convergence interval (I can use the Ratio test here, I believe) but how would I find the Maclaurin series for that using the one given?
$$\frac{1}{1-x}=\sum^\infty_{n=0}x^n$$
$\endgroup$2 Answers
$\begingroup$$$ \dfrac{1+x}{1+x^2} = (1+x)\dfrac1{1+x^2}=(1+x)\dfrac1{1-(-x^2)}$$
$\endgroup$ $\begingroup$We obtain: \begin{align*} \color{blue}{\frac{1+x}{1+x^2}} &=(1+x)\sum_{j=0}^\infty(-x^2)^j\tag{1}\\ &=\sum_{j=0}^\infty(-1)^jx^{2j}+\sum_{j=0}^\infty(-1)^jx^{2j+1}\tag{2}\\ &=\sum_{j=0}^\infty(-1)^{\left\lfloor\frac{j}{2}\right\rfloor} x^{j}\tag{3}\\ &\,\,\color{blue}{=1+x-x^2-x^3+x^4+x^5-\cdots} \end{align*}
Comment:
In (1) we use the geometric series expansion.
In (2) we multiply out.
In (3) we use the floor function $\lfloor x\rfloor$ to determine the correct sign.
