0
$\begingroup$

I want to kmow what series the function $1/(1-ax)^r, a,r\in N $ generats. I thoghut about doing this:

lets name y=ax now we have $1/(1-y)^r, r\in N $ and we know $1/(1-y)^r= \sum_{n=0}^{\infty}{n+r-1\choose r-1}y^n$

now lets put back y=ax so $1/(1-ax)^r = \sum_{n=0}^{\infty}{n+r-1\choose r-1}a^nx^n$

does is make sense?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ how did u do the second transition ? from (-r choose k) to (r+k-1 choose k ) ? $\endgroup$ – KIMKES1232 Jun 12 at 16:30
  • $\begingroup$ 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 ${-4 \choose 3}-=-20$ $\endgroup$ – Dr Zafar Ahmed DSc Jun 12 at 17:37
  • $\begingroup$ @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\ $\endgroup$ – Dr Zafar Ahmed DSc Jun 12 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.