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It's said that the generating function for $g(x) = \sum_{d=0}^\infty {d+m-1 \choose m-1} x^d$ is equal to $\frac{1}{(1-x)^m}$.

In the proof that I have seen it states that:

By the geometric series, $\frac{1}{1-x} = 1+x+x^2+... = \sum_{d=0}^\infty {d \choose 0} x^d$

Differentiating gives $\frac{1}{(1-x)^2} = \sum_{d=0}^\infty {d \choose 1} x^d$

Differentiating again gives $\frac{1}{(1-x)^3} = \sum_{d=0}^\infty {d \choose 2} x^d$

And so on, but does this not show that the generating function is $g(x) = \sum_{d=0}^\infty {d \choose m-1} x^d$ or am I missing something?

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  • $\begingroup$ $\sum_{0}^{\infty}$ is ambiguous. You probably mean $\sum_{d=0}^{\infty}$ $\endgroup$ – Alex Apr 22 '16 at 12:56
  • $\begingroup$ $\binom{j}{k}$ for $j<k$ is 0 $\endgroup$ – Alex Apr 22 '16 at 12:57
  • $\begingroup$ Differentiating should give $$\frac{1}{(1-x)^2}=\sum_{d} \binom{d+1}{1}x^d$$ The +1 is very important. $\endgroup$ – Thomas Andrews Apr 22 '16 at 13:36
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The first part requires a proof by induction that is merely hinted at in what you’ve written: if

$$\frac1{(1-x)^m}=\sum_{d\ge 0}\binom{d+m-1}{m-1}x^d\;,$$

differentiating yields

$$\begin{align*} \frac{m}{(1-x)^{m+1}}&=\sum_{d\ge 0}d\binom{d+m-1}{m-1}x^{d-1}\\ &=\sum_{d\ge 0}d\binom{d+m-1}dx^{d-1}\\ &=\sum_{d\ge 0}(d+m-1)\binom{d+m-2}{d-1}x^{d-1}\\ &=\sum_{d\ge 0}(d+m)\binom{d+m-1}dx^d\\ &=\sum_{d\ge 0}(d+m)\binom{d+m-1}{m-1}x^d\\ &=\sum_{d\ge 0}m\binom{d+m}mx^d\;. \end{align*}$$

Now just divide both sides by $m$ to get

$$\frac1{(1-x)^{m+1}}=\sum_{d\ge 0}\binom{d+m}mx^d\;.$$

You’re right that this doesn’t justify the $g(x)$ that you’ve written. In fact

$$\sum_{d\ge 0}\binom{d+m}mx^d=\sum_{d\ge 0}\binom{d}mx^{d-m}\;,$$

since $\binom{k}\ell=0$ if $k<\ell$, and therefore

$$\sum_{d\ge 0}\binom{d}mx^d=\frac{x^m}{(1-x)^{m+1}}\;.$$

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  • $\begingroup$ Perfect, thank you so much! $\endgroup$ – user328251 Apr 22 '16 at 13:26
  • $\begingroup$ @user328251: You’re welcome! $\endgroup$ – Brian M. Scott Apr 22 '16 at 13:27

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