# 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)?

• This is S(1+x) where $S=1+2x^2+3x^4+...$ for a start. – Paul Dec 19 '14 at 23:18
• @paul: What about x? – Mathy Person Dec 19 '14 at 23:21
• @Paul Saying $(1+x)S(x)$ might be more clear. – Tim Raczkowski Dec 19 '14 at 23:25

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: $$= (1+x) \frac{1}{(1-y)^2}$$ $$= \frac{1+x}{(1-x^2)^2}$$ $$= \frac{1}{(1-x)(1+x^2)}.$$

• I believe $(y+y^2+y^3+y^4+....)$ = $\frac{1}{1-y}$, if I am not mistaken. Then is it: $\frac{1+x}{1-y}$? – Mathy Person Dec 19 '14 at 23:28
• Oh wait, I see that you had $\frac{y}{1-y}$ instead. Is it: $\frac{(1+x)(y)}{1-y}$? – Mathy Person Dec 19 '14 at 23:30
• And $\frac{(1+x)(y)}{1-y}$, and $y=x^2$, then $\frac{(1+x)(x^2)}{1-x^2}$? – Mathy Person Dec 19 '14 at 23:30
• Simplifying would result in: $\frac{x^2}{1-x}$? – Mathy Person Dec 19 '14 at 23:31
• Just take derivative in y of $y/(1-y)$, then replace $y$ with $x^2$. – ir7 Dec 19 '14 at 23:33

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 $-1 < x - 1$.

• Hmm..I got $\frac{x^2}{1-x}$ (see what I posted in reply to ir7's comment). Did I make a mistake somewhere? – Mathy Person Dec 19 '14 at 23:33
• Whoops my bad, I forgot a step – SBareS Dec 19 '14 at 23:36
• Apparently I am to tired for maths right now... – SBareS Dec 20 '14 at 0:05

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.

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}