Expand $(1-z)^{-m}$, $m$ a positive integer, in powers of $z$.
Since $\dfrac{1}{1-z}=1+z+z^2+\ldots$, we can find
$$\dfrac{1}{(1-z)^2} = (1+z+z^2+\ldots)(1+z+z^2+\ldots) = 1+2z+3z^2+\ldots.$$
So, in a similar way, by using induction and multiplying by $1+z+z^2+\ldots$ at each step, we can find
$$\dfrac{1}{(1-z)^n} = 1+\binom{n}{n-1}z+\binom{n+1}{n-1}z^2+\binom{n+2}{n-1}z^3+\ldots.$$
I wonder whether this is a rigorous solution to the question or not. If not, what needs to be changed?
$\endgroup$ 24 Answers
$\begingroup$As written this isn't really rigorous. You hint at using induction to prove this, so you should actually demonstrate (by writing down an induction argument) that it works, instead of just showing it for a few cases.
That is, suppose that $$\frac{1}{(1-z)^n} = 1 + \binom{n}{n-1}z + \binom{n+1}{n-1}z^2 + \binom{n+2}{n-1}z^3 + \cdots$$ and then derive from this that $$\frac{1}{(1-z)^{n+1}} = 1 + \binom{n+1}{n}z + \binom{n+2}{n}z^2 + \binom{n+3}{n}z^3 + \cdots$$
You should also stipulate that this is only valid when $\left| z \right| < 1$.
$\endgroup$ $\begingroup$Hint: You can use Taylor series of a function at $z=0$
$$ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n. $$
Note that,
$$ f(z)=(1-z)^{-m} \implies f^{(n)}(z)= (m)(m+1)\dots(m+n-1)(1-z)^{-m-n} $$
You can simplify the above product using Pochhammer symbol.
$\endgroup$ $\begingroup$HINT: Use the fact that $$\frac{\operatorname{d}^k}{\operatorname{d}\!z^k}\left(\frac{1}{1-z}\right) = \frac{k!}{(1-z)^{k+1}}$$
$\endgroup$ $\begingroup$You want to show that given your proposed formula works for case $n$, then it works for case $n+1$. Combining this with showing that the "base case" works, then this will be a proper induction argument (and this is the general structure of all induction proofs).
As an aside, another way you could try to prove the identity is by using the fact that $$ \frac{d}{dz} (1 - z)^{-n} = n (1 - z)^{-n-1}$$ and then writing down what the recursive formula for the coefficients have to be.
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