Differential calculus prove of $\sum_{i=0}^n {{i}\choose{ k}} = {{n+1}\choose{ k+1}}$

Question

I asked in previous question for combinatorial proof of $\sum_{i=0}^n {{i}\choose{ k}} = {{n+1}\choose{ k+1}}$ but I heard it's possible to prove it using differential calculus (for sums). How to do that?

Answer 1

$$ \begin{align} \sum_{k=0}^n\sum_{j=0}^n\binom{j}{k}x^k &=\sum_{j=0}^n(1+x)^j\tag{1}\\ &=\frac{(1+x)^{n+1}-1}{x}\tag{2}\\[3pt] &=\sum_{k=0}^n\binom{n+1}{k+1}x^{k}\tag{3} \end{align} $$ Explanation:
$(1)$: Binomial Theorem
$(2)$: Sum of a Geometric Series
$(3)$: Binomial Theorem

Equate the powers of $x$.

Answer 2

Let $\mathbb{D}$ be the open unit disc $\big\{z\in\mathbb{C}\,\big|\,|z|<1\big\}$ in $\mathbb{C}$. Note that $(1-z)^{-k-1}=\displaystyle\sum_{r=0}^\infty\,\binom{r+k}{k}\,z^r$ for $z\in\mathbb{D}$. Consider $f(z):=\displaystyle\sum_{j=0}^{n-k}\,\frac{(1-z)^{-k-1}}{z^{j+1}}$ for all $z\in \mathbb{D}\setminus\{0\}$.

Let $\gamma$ be a closed smooth curve in $\mathbb{D}$ that encircles the origin once in the counterclockwise direction. Then, $$\frac{1}{2\pi\text{i}}\,\oint_\gamma\,f(z)\,\text{d}z=\sum_{j=0}^{n-k}\,\binom{j+k}{k}=\sum_{i=k}^n\,\binom{i}{k}\,,$$
as $\displaystyle\frac{1}{2\pi\text{i}}\,\oint_\gamma\,\frac{(1-z)^{-k-1}}{z^{j+1}}\,\text{d}z=\binom{j+k}{k}$ for all $j=0,1,2,\ldots,n$.

On the other hand, for any $z\in\mathbb{D}\setminus\{0\}$, we have
$$\begin{align} f(z)&=(1-z)^{-k-1}\,\sum_{j=0}^{n-k}\,\frac{1}{z^{j+1}}=\frac{(1-z)^{-k-1}}{z}\left(\frac{1-\frac{1}{z^{n-k+1}}}{1-\frac{1}{z}}\right) \\ &=\frac{(1-z)^{-k-2}\left(1-z^{n-k+1}\right)}{z^{n-k+1}}=\frac{(1-z)^{-k-2}}{z^{n-k+1}}-(1-z)^{-k-2}\,. \end{align}$$ Hence, $\displaystyle\frac{1}{2\pi\text{i}}\,\oint_{\gamma}\,f(z)\,\text{d}z=\frac{1}{2\pi\text{i}}\,\oint_{\gamma}\,\frac{(1-z)^{-k-2}}{z^{n-k+1}}\,\text{d}z$ because $\displaystyle\oint_\gamma\,(1-z)^{-k-2}\,\text{d}z=0$. As $$(1-z)^{-k-2}=\displaystyle\sum_{r=0}^\infty\,\binom{r+k+1}{k+1}\,z^r$$ for all $z\in\mathbb{D}$, we obtain $$\frac{1}{2\pi\text{i}}\,\oint_\gamma\,f(z)\,\text{d}z=\binom{(n-k)+k+1}{k+1}=\binom{n+1}{k+1}\,.$$

Answer 3

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\begin{align} \sum_{i = 0}^{n}{i \choose k} & = \sum_{i = 0}^{n}\ \overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{i} \over z^{k + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{=\ {i \choose k}}}\ =\ \oint_{\verts{z} = 1}{1 \over z^{k + 1}}\sum_{i = 0}^{n}\pars{1 + z}^{i} \,{\dd z \over 2\pi\ic} \\[3mm] & = \oint_{\verts{z} = 1}{1 \over z^{k + 1}} {\pars{1 + z}^{n + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}\ =\ \overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{n + 1} \over z^{k + 2}} \,{\dd z \over 2\pi\ic}}^{\ds{=\ {n + 1 \choose k + 1}}}\ -\ \overbrace{\oint_{\verts{z} = 1}{1 \over z^{k + 2}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \delta_{k,-1}}} \\[3mm] & = \fbox{$\ds{{n + 1 \choose k + 1} - \delta_{k,-1}}$} \end{align}