if $\sum_0^\infty a_n x^n = (\sum_0^\infty x^n )(\sum_0^\infty x^{2n})$ what is $a_n$? if $$\sum_0^\infty a_n x^n = (\sum_0^\infty x^n )(\sum_0^\infty x^{2n})$$
what is $a_n$?
Here is my approach 
let $b_n= 1$ and $c_n= x^n$
Then by forming / relating to cauchy product we can conlude that the product is equal to : 
$$\sum_0^\infty a_n x^n$$
where $a_n = \sum_{k=0}^{n} b_k c_{n-k} = \sum x^{n-k}= x^n+x^{n-1}+...+x^0 = \frac{1-x^{n+1}}{1-x}$
What do you think about my approach? I feel like there is something wrong with it. If not - could you provide a different approach? By the way depending on the way I solve - I keep finding different answers. I also ended up with $a_n=\frac{3+2n+(-1)^n}{4}$ without using cauchy product.
 A: Using the geometric summation formula $\displaystyle\sum x^n=\dfrac1{1-x}$,
$$A=\frac1{1-x}\frac1{1-x^2}=\frac1{4(1+x)}+\frac1{4(1-x)}+\frac1{2(1-x)^2}\\
=\frac1{4(1+x)}+\frac1{4(1-x)}+\frac12\frac d{dx}\frac x{1-x}$$ so that
$$a_n=\frac14(-1)^n+\frac14+\frac12(n+1).$$
This is a generating function approach.
A: Let us write the first sum on the RHS as a geometrical series:
$$\sum_{n=0}^\infty a_n x^n =\dfrac{1}{1-x}\sum_{n=0}^\infty x^{2n} \ \ \iff$$
$$(1-x)\sum_{k=0}^\infty a_k x^k =\sum_{n=0}^\infty x^{2n} \ \ \iff$$
$$\tag{1}\sum_{k=0}^\infty a_{k}x^k - \sum_{k=0}^\infty a_{k}x^{k+1} = \sum_{n=0}^\infty x^{2n}.$$
Setting $K:=k+1 \ \iff \ k=K-1$, (1) becomes:
$$\sum_{k=0}^\infty a_{k}x^k - \sum_{K=1}^\infty a_{K-1}x^{K} = \sum_{n=0}^\infty x^{2n}$$
$$a_0+\sum_{k=1}^\infty (a_{k}-a_{k-1}) x^k =\sum_0^\infty x^{2n} $$
An immediate consequence is that $a_0=1$ ; then, for positive indices, identitifying the coefficients in the LHS and RHS, two cases occur, depending if they correspond to odd or even exponents:
$$\begin{cases}k&=&2n+1& \ \implies \ & a_{2n+1}-a_{2n}=0\\k&=&2n& \ \implies \ & a_{2n}-a_{2n-1}=1 \end{cases}$$
In other words: $$\begin{cases}a_{2n+1}=a_{2n}\\a_{2n}=a_{2n-1}+1 \end{cases}.$$
Summarizing, we have a unit increase at even steps, and no increase at odd steps.
Hence, taking into account value $a_0=1$, we get the following explicit formula for the general term:

$$a_n=\left\lfloor \frac{n}{2} \right\rfloor+1$$

A: Generally 
$$\sum_0^\infty c_n x^n = (\sum_0^\infty a_n x^n )(\sum_0^\infty b_nx^{n})$$
Then define $b_{2k+1} =0$ 
$$\sum_0^\infty c_n x^n = (\sum_0^\infty a_n x^n )(\sum_0^\infty b_{2k}x^{2k})$$
Then you can use the Cauhcy product formula.

Now for the Cauchy product 
$$\sum_0^\infty c_n x^n = (\sum_0^\infty a_n x^n )(\sum_0^\infty b_nx^{n})$$
Let $a_n=1$ and $b_{2k}=1,b_{2k+1} = 0$, we conclude 
$$c_n = \sum_{k=0}^{n} b_k a_{n-k} = \sum_{k=0}^nb_k $$
Let us see the case for even $n = 2j$
$$\sum_{k=0}^{2j} b_k =\sum_{k=0}^{j} b_{2k} +\sum_{k=0}^{j-1} b_{2k+1} = \sum_{k=0}^{j} 1 = j + 1 = \frac{n}{2} +1 $$
Let us see the case for odd $n = 2j +1$
$$\sum_{k=0}^{2j+1} b_k =\sum_{k=0}^{j} b_{2k} +\sum_{k=0}^{j} b_{2k+1} = \sum_{k=0}^{j} 1 = j + 1 = \frac{n-1}{2} +1 $$
We deduce that 
$$\left(\sum_{n=0}^\infty x^n \right) \left(\sum_{n=0}^\infty x^{2n} \right) = \sum_{n=0}^\infty \left(\left\lfloor \frac{n}{2} \right\rfloor+1 \right)x^n$$
Now since $$\left\lfloor \frac{n}{2} \right\rfloor+1 =\frac{3+(-1)^n+2n}{4}$$
We conclude that 

$$\left(\sum_{n=0}^\infty x^n \right) \left(\sum_{n=0}^\infty x^{2n}
 \right) = \sum_{n=0}^\infty \left(\frac{(3+(-1)^n+2n}{4} \right)x^n$$


Another example
Calculate the series expansion at $x=0$ of the integral $\int \frac{xy\arctan(xy)}{1-xy}dx$
A: HINT: the statement for convergent series
$$\sum_{k=0}^\infty a_k x^k=\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=0}^\infty x^{2k}\right)$$ implies that
$$\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=0}^\infty x^{2k}\right)=\left(\lim_{n\to\infty}\sum_{k=0}^n x^k\right)\left(\lim_{m\to\infty}\sum_{k=0}^m x^{2k}\right)$$
In particular we have the case
$$\begin{align}\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=0}^\infty x^{2k}\right)&=\left(\lim_{n\to\infty}\sum_{k=0}^{2n} x^k\right)\left(\lim_{n\to\infty}\sum_{k=0}^n x^{2k}\right)\\&=\lim_{n\to\infty}\left(\sum_{k=0}^{2n}x^k\right)\left(\sum_{k=0}^n x^{2k}\right)\\&=\lim_{n\to\infty}\left(\sum_{k=0}^{2n}x^k\right)\left(\sum_{k=0}^{2n} x^{k}\chi_{2\Bbb N}(k)\right)\end{align}$$
provided that each series converges. From here is easy to use the Cauchy product to define the $a_k$ where the function $\chi_{2\Bbb N}$ is defined as
$$\chi_{2\Bbb N}(k):=\begin{cases}1,& k\in 2\Bbb N_{\ge 0}\\0,&k\notin 2\Bbb N_{\ge 0}\end{cases}$$
