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$f(x) = \frac{x}{6x^2-5x+1}$ is the generating function of a series. How do I go about finding the formula of this series?

I started by finding the partial fractions: $$f(x) = \frac{1}{2x-1} + \frac{-1}{3x-1}$$ How do I proceed from here?

Etemon
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bosendorfer
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    Hint: Use $$\frac{1}{1-x}=1+x+x^2+...$$ – Leonard Neon Jan 08 '21 at 18:43
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    Once your denominators are linear terms, you can use the fact that $\frac{1}{1-r} = 1+r+r^2+\cdots$. So for example in this case, $\frac{1}{2x-1} = -\frac{1}{1-2x} = -(1+2x+(2x)^2+\cdots)$ and $-\frac{1}{3x-1} = \frac{1}{1-3x} = 1+3x+(3x)^2+\cdots$. – Herman Chau Jan 08 '21 at 18:44

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hint

As recomended, by Leon,

If $ |X|<1 $, then

$$\frac {1}{1-X}=1+X+X^2+...=\sum_{k=0}^{+\infty}X^k$$

So, if $ |3x|<1 $,

$$f(x)=\frac{1}{1-3x}-\frac{1}{1-2x}=$$

$$\sum_{k=0}^{+\infty}((3x)^k-(2x)^k)$$

hamam_Abdallah
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  • Now, correct me if I'm wrong, but would that mean, that the series I'm looking for is $a_{k} = \frac{(3x)^k - (2x)^k}{x^k}$ ? – bosendorfer Jan 08 '21 at 19:12
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    @bosendorfer It is $ \sum (3^k-2^k)x^k$ – hamam_Abdallah Jan 08 '21 at 19:13
  • How so? According to Wikipedia generating function formula is: $$\sum_{n=0}^{\infty}a_{n}\cdot x^n$$ So my generating function $f(x)=\sum_{k=0}^{+\infty} (3x)^k - (2x)^k$ doesn't exactly fit the Wikipedia's formula for generating function. Until, that is, I change the $(3x)^k-(2x)^k$ to $\frac{(3x)^k-(2x)^k}{x^k} \cdot x^k$. How is my thinking wrong? – bosendorfer Jan 08 '21 at 19:24
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    Here, your $ a_n $ will be simply $ 3^n-2^n$. – hamam_Abdallah Jan 08 '21 at 19:26
  • Oh yes, I see it now. Thank you. I have one more question regarding your initial answer: if $|3x|\geq 1$ would it be possible to solve this task? – bosendorfer Jan 08 '21 at 19:32
  • @bosendorfer This condition is equivalent $ -1/3<x<1/3$. I gives the Radius of convergence of the power series. – hamam_Abdallah Jan 08 '21 at 19:36