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.

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 $\int \frac{xy\arctan(xy)}{1-xy}dx$
