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I rewrote the equation series as $$ \frac69 \sum_{n=0}^\infty \left(\frac{-1}{9}\right)^n x^{n+1} $$

And therefore have coefficients of $C_0 = 6/9, C_1 = \left( 6/9 \right) \left(\frac{-1}{9}\right)^1, C_2 = ( 6/9 )\left(\frac{-1}{9}\right)^2, C_3 = \cdots $ This is wrong however and I'm not sure why. If anyone knows the solution to this could they point me in the right direction?

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    $\begingroup$ Idea in general is right, but for example $C_0=0$. $\endgroup$ – André Nicolas Aug 9 '14 at 2:33
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    $\begingroup$ Rewrite $\frac{6x}{x+9}$ as $\frac{6x}{9}\cdot \frac{1}{1+\frac{x}{9}}$. Expand the last term, you know how to do that. Multiply by $\frac{6x}{9}$. We get $C_0=0$, $C_1=\frac{6}{9}=\frac{2}{3}$, $C_2=-\frac{2}{27}$, and so on. $\endgroup$ – André Nicolas Aug 9 '14 at 2:39
  • $\begingroup$ Totally get it! Thanks! $\endgroup$ – BLZN Aug 9 '14 at 2:48
  • $\begingroup$ You are welcome, you basically had it already. $\endgroup$ – André Nicolas Aug 9 '14 at 2:51
  • $\begingroup$ "Equation" is of course the wrong word here. $\endgroup$ – Michael Hardy Aug 9 '14 at 3:17
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$\frac{6x}{x+9}=6-\frac{54}{9+x}=6-\frac{6}{1-\frac{-x}{9}}=6-6\sum_{n=0}^{\infty}(-\frac{x}{9})^n \quad \forall x,\mid{x}\mid<9$.

Now let $n=0,1,2,4$ and calculate $c_{n}$s.

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For sure, the fastest way is to build the Taylor series of the lhs and identify the coefficients.

You could also write $$6x =(x+9) \sum_{n=0}^4 C_n x^n=9 C_0+(C_0+9 C_1) x+(C_1+9 C_2) x^2+(C_2+9 C_3) x^3+(C_3+9 C_4) x^4+C_4 x^5+ ...$$ and identify the coefficients for each power of $x$. This gives $C_0=0$, $9C_1=6$ and so on.

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