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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)?

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

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

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

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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 $1+x$: $$ \begin{align} 1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+\dots &=\frac{1+x}{(1-x^2)^2}\\ &=\frac1{(1-x)(1-x^2)} \end{align} $$

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