How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)? 
 A: Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots) $$
$$ =(1+x)(1+2y+3y^2+4y^3 +\ldots)$$
$$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$
$$ = (1+x)\left( \frac{y}{1-y}\right)'$$
Edit:
$$ = (1+x) \frac{1}{(1-y)^2} $$
$$ = \frac{1+x}{(1-x^2)^2}  $$
$$ = \frac{1}{(1-x)(1+x^2)}.$$
A: I would go about this by first splitting the series up:
$$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$
Letting $s=1+2x^2+3x^4$ we can do a few tricks:
$$s-x^2s=\begin{array}{c} 1&+2x^2&+3x^4+... \\ &-x^2&-2x^4-...\end{array}$$
$$=1+x^2+x^4+...$$
Which converges to $\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving this is not hard, and can be done by a technique like the above). This gives 
$$s -x^2s=\frac{1}{1-x^2}\Leftrightarrow s=\frac{1}{(1-x^2)^2}=$$
Thus the original series converges to:
$$(1+x)s=\frac{(1+x)}{(1-x^2)^2}$$
For $-1 < x - 1$.
A: Unless I'm mistaken, it is $$\sum_{n=1}^\infty nx^{2n-2} + \sum_{n=1}^\infty nx^{2n-1}$$
If you can compute one of the two terms, e.g. $\sum_{n=1}^\infty nx^{2n-2} = \sum_{n=0}^\infty (n+1)x^{2n} = \sum_{n=0}^\infty (n+1)(x^{2})^n$ (see e.g. this, then you'll also get the other term (by multiplying it by $x$), and thus the sum.
A: One way is to look at 
$$
1+2x^2+3x^4+4x^6+5x^8+\dots
$$
as
$$
1+2t+3t^2+4t^3+5t^4+\dots
$$
where $t=x^2$. The last series is the derivative of
$$
1+t+t^2+t^3+t^4+t^5+\dots=\frac1{1-t}
$$
Therefore,
$$
1+2t+3t^2+4t^3+5t^4+\dots=\frac1{(1-t)^2}
$$
and
$$
1+2x^2+3x^4+4x^6+5x^8+\dots=\frac1{(1-x^2)^2}
$$
Now, just multiply by $1+x$:
$$
\begin{align}
1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+\dots
&=\frac{1+x}{(1-x^2)^2}\\
&=\frac1{(1-x)(1-x^2)}
\end{align}
$$
