It is easily checked that
$\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0$, for example by appealing to the binomial theorem.
I'm trying to figure out what happens with the truncated sum
$\displaystyle\sum_{i\ =\ 0}^{D}\left(\, -1\,\right)^{i}\binom{n}{i}$.
How far away from $0$ can this get, as a function of $D$ ?.
I'm mostly interested in the case of when $D \ll n$, such as
$D \sim \,\sqrt{\,n\,}\,$.
Thanks !
Best Answer
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$\begin{align} &\color{#88f}{\large\sum_{k = 0}^{D}\pars{-1}^{k}{n \choose k}} =\sum_{k = 0}^{D}\pars{-1}^{k}\ \overbrace{\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1} {\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ {n \choose k}}} \\[5mm]& \ =\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{n} \over z} \sum_{k = 0}^{D}\pars{-\,{1 \over z}}^{k}\,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{n} \over z} {\pars{-1/z}^{D} + z \over 1 + z}\,{\dd z \over 2\pi\ic} \\[5mm]&=\pars{-1}^{D}\ \underbrace{\oint_{\verts{z}\ =\ a\ <\ 1} {\pars{1 + z}^{n - 1} \over z^{D + 1}}\,{\dd z \over 2\pi\ic}} _{\ds{=\ {n - 1 \choose D}}}\ +\ \underbrace{\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}\pars{1 + z}^{n - 1} \,{\dd z \over 2\pi\ic}}_{\ds{=\ 0}} \\[5mm]&\ = \bbox[10px,border:1px groove navy]{\pars{-1}^{D}{n - 1 \choose D}} \\ & \end{align}