Find 'ordinary generating function': 1, 0, 2, 0, 3, 0, 4, 0, 5....

Question

This was on my midterm yesterday, but I couldn't solve it.

Find an ordinary generating function: 1, 0, 2, 0, 3, 0, 4, 0, 5....

My answer was $(\frac{1}{1-x^2})^2$ because 1, 2, 3, 4, 5, 6, 7, ... is $\frac{1}{1-x^2}$. (now I feel like it just doesn't make sense).

Can you find the ordinary generating function?

Answer 1

We obtain \begin{align*} \color{blue}{1+2x^2+3x^4+\cdots}&=\sum_{n=0}^\infty (n+1)x^{2n}\\ &=\frac{1}{2x}\frac{d}{dx}\left(\sum_{n=0}^\infty \left(x^2\right)^{n+1}\right)\\ &=\frac{1}{2x}\frac{d}{dx}\left(\sum_{n=1}^\infty \left(x^2\right)^n\right)\\ &=\frac{1}{2x}\frac{d}{dx}\left(\frac{1}{1-x^2}-1\right)\\ &\,\,\color{blue}{=\frac{1}{(1-x^2)^2}} \end{align*}

Answer 2

Observe that the sequence will correspond to the generating function $$ (1+x^2+x^4+x^6\dotsb)+(x^2+x^4+x^6\dotsb)+(x^4+x^6+\dotsb)=\sum_{k=0}^\infty\frac{x^{2k}}{1-x^2}=\frac{1}{(1-x^2)^2} $$ since $$ \sum_{k=0}^\infty x^{2k}=\frac{1}{1-x^2} $$ The manipulations above make sense as identities of analytic functions for $|x|<1$ or as identities of formal power series.