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I want to know what series the function $$1/(1-ax)^r, \quad a,r\in \mathbb{N}, $$ generates.

I thought about doing this: Let's name $y=ax$. Now we have $$\frac{1}{(1-y)^r}, \quad r\in \mathbb{N},$$ and we know that $$\frac{1}{(1-y)^r}= \sum_{n=0}^{\infty}{n+r-1\choose r-1}y^n.$$ Now let's put back $y=ax$. So $$\frac{1}{(1-ax)^r} = \sum_{n=0}^{\infty}{n+r-1\choose r-1}a^nx^n.$$ Does this make sense?

Mars Plastic
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KIMKES1232
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1 Answers1

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The series is $$S=(1-ax)^{-r}=\sum_{k=0}^{\infty}(-1)^k {-r \choose k} (ax)^k=\sum_{k=0}^{\infty} {r+k-1 \choose r - 1} (ax)^k.$$ Which is valid for $|x|<a^{-1}.$

vonbrand
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Z Ahmed
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  • how did u do the second transition ? from (-r choose k) to (r+k-1 choose k ) ? – KIMKES1232 Jun 12 '19 at 16:30
  • Good question, note that ${s \choose k}=\frac{s(s-1)(s-2)(s-3)....(s-k+1)}{k!}$, in this s may not be positive integer. It can be negative integer or non-integer or even complex. Check that ${-1 \choose k}=(-1)^k$, ${-3 \choose 2}=-3(-3-1)/2$ etc. You can check the identity thar ${-r \choose k}=(-1)^k { r+k-1 \choose k.}$ Check ${-4 \choose 3}-=-20$ – Z Ahmed Jun 12 '19 at 17:37
  • @KIMKES1232 One more interesting way to calculate ${-4 \choose 3}=\frac{(-4)!}{(-7)! 3!}.$ In this $(-4)!$= product of all integers from $-\infty$ to -4. Similarly $(-7)!$ means product of all numbers from $-\infty$ up to $-7$. do the cancellations to get -20\ – Z Ahmed Jun 12 '19 at 18:16