What is $a_n$, if $\sum_0^\infty a_n x^n = (\sum_0^\infty x^n )(\sum_0^\infty x^{2n})$ I found a answer here https://math.stackexchange.com/a/2174836/533661
But why the answer uses $\sum x^n=\frac{1}{1-x}$? To use that, we must assume $|x|<1$, but the question does not assume that.
And how the answer got $a_n$ from $a_{2n+1}=a_{2n}, a_{2n}=a_{2n-1}+1$
 A: \begin{eqnarray*}
\sum_{n=0}^{\infty} x^n \sum_{m=0}^{\infty} x^{2m} = \frac{1}{(1-x)} \frac{1}{(1-x^2)} 
\end{eqnarray*}
Multiply top & bottom by $(1+x)$ and split the function into its odd & even powers.
\begin{eqnarray*}
\frac{1+x}{(1-x^2)^2} &=& \color{blue}{\frac{1}{(1-x^2)^2}} +\frac{x}{(1-x^2)^2} \\
&=& \color{blue}{\sum_{i=0}^{\infty} (i+1) x^{2i}} +\sum_{i=0}^{\infty} (i+1) x^{2i+1}.
\end{eqnarray*}
The above series can be considered as "Formal Power series" to represent the sequence $1,1,2,2,3,3,\cdots$ and for combinatorialists their convergence in an analytic sense is not of interest.
A: hint
For the second point:
If we put $$b_n=a_{2n+1}$$
then
$$b_n=a_{2n}=a_{2n-1}+1=b_{n-1}+1$$
thus $ (b_n) $ is arithmetic. So,
$$b_n=b_0+n=a_1+n=a_{2n+1}$$
and
$$a_{2n}=a_{2n+1}=a_1+n$$
A: Well, we could write it this way:
$$\begin{aligned} \sum_{n=0}^\infty a_n x^n &= \left(\sum_{j=0}^\infty x^j \right)\left(\sum_{k=0}^\infty x^{2k}\right)\\
& =\sum_{j=0}^\infty \sum_{k=0}^\infty x^j x^{2k} \\
& =\sum_{j=0}^\infty \sum_{k=0}^\infty x^{j+2k} \\
\end{aligned} 
$$
but if we notice that
$$\sum_{x=0}^\infty  x^n=\frac{1}{1-x}$$
and
$$\sum_{x=0}^\infty  x^{2n}=\frac{1}{1-x^2}$$
so that
$$\sum_{n=0}^\infty a_n x^n = \frac{1}{1-x} \cdot \frac{1}{1-x^2}$$
we could make use of the coefficients for the McLauren series.  Better yet, write this as a partial fraction expansion (as was done over at the link to the solution) and read off the coefficients.
Both series converge if and only if $|x|<1$, so that is understood.
