Why does $\sum_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$? I don't understand the identity $\sum_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$, where $k$ is fixed.
I first approached it by considering the identity
$$
\sum_{n,k\geq 0} \binom{n}{k} x^n y^k = \sum_{n=0}^\infty x^n \sum_{k=0}^n \binom{n}{k} y^k = \sum_{n=0}^\infty x^n (1+y)^n  = \frac{1}{1-x(1+y)}.
$$
So setting $y=1$, shows $\sum_{n,k\geq 0}\binom{n}{k}x^n=\frac{1}{1-2x}$. What happens if I fix some $k$ and let the sum range over just $n$? Thank you.
 A: Call $s_k(x)=\sum\limits_n{n\choose k}x^k$. You know that $\sum\limits_ks_k(x)y^k=t(x,y)$ with $t(x,y)=\frac1{1-x(1+y)}$. Hence $s_k(x)$ is the coefficient of $y^k$ in the series expansion of $t(x,y)$ as a function of $y$, which we now compute.
Note that $t(x,y)=\frac1{1-x}\frac1{1-r(x)y}$ with $r(x)=\frac{x}{1-x}$ and that the series expansion of $\frac1{1-z}$ is $\sum\limits_kz^k$. Using this for $z=r(x)y$, one gets $t(x,y)=\frac1{1-x}\sum\limits_kr(x)^ky^k$. By identification of the coefficient of $y^k$, one sees that $s_k(x)=\frac1{1-x}r(x)^k$.
A: You can work directly with properties of the binomial coefficient. For $k\ge 0$ let $$f_k(x)=\sum_{n\ge 0}\binom{n}kx^n\;.$$ Then
$$\begin{align*}
f_k(x)&=\sum_{n\ge 0}\binom{n}{k}x^n\\
&=\sum_{n\ge 0}\left[\binom{n-1}{k-1}+\binom{n-1}{k}\right]x^n\\
&=\sum_{n\ge 0}\left[\binom{n}{k-1}+\binom{n}k\right]x^{n+1}\\
&=x\sum_{n\ge 0}\binom{n}{k-1}x^n+x\sum_{n\ge 0}\binom{n}kx^n\\
&=xf_{k-1}(x)+xf_k(x)\;,
\end{align*}$$
so $(1-x)f_k(x)=xf_{k-1}(x)$, and $$f_k(x)=\frac{x}{1-x}f_{k-1}(x)\;.\tag{1}$$
Since $$f_0(x)=\sum_{n\ge 0}\binom{n}0x^n=\sum_{n\ge 0}x^n=\frac1{1-x}\;,$$ an easy induction yields the desired result.
A: Start from the Taylor series of $\frac1{1-x}$, differentiate it $k$ times, and then multiply everything by $\frac{x^k}{k!}$.
A: The easiest way to see this is to use the common generalization of binomial coefficients with negative upper index, which satisfy
$$
  \binom{-n}k=(-1)^k\binom{n+k-1}k,
$$
and for which the binomial theorem in the form
$$
  (1+X)^{-n}=\sum_{k\geq0}\binom{-n}kX^k
$$
continues to hold, as an identity of formal power series.
Now in order to get your exponents to match the lower index, write
$$\begin{align}
  \sum_{n\geq k}\binom nkX^n
&=\sum_{n\geq k}\binom n{n-k}X^n
=\sum_{m\geq0}\binom{k+m}mX^{k+m}
\\&
=X^k\sum_{m\geq0}(-1)^m\binom{-k-1}mX^m
\quad\text{by above identity}
\\&
=X^k(1-X)^{-k-1}
\quad\text{by the binomial theorem}
\\&
=\frac{X^k}{(1-X)^{k+1}},
\end{align}
$$
an identity of formal power series (which continues to hold when substituting a value for $X$ small enough to make the series on the left converge).
