Finite Sum evaluation

Question

Can anyone help me in proving the following identity: For any $k\in\mathbb{N}$, $$ \sum_{n=0}^{k}\frac{(-1)^n(k-n)^{2k}}{n!(2k-n)!}=\frac{1}{2}. $$ Thank You very much in advance.

Answer 1

Note that by letting $m=2k-n$ we get $$\sum_{n=0}^{k}\frac{(-1)^n(k-n)^{2k}}{n!(2k-n)!}=\sum_{m=k}^{2k}\frac{(-1)^{m}(k-m)^{2k}}{(2k-m)!(m)!}$$ Hence $$\frac{1}{(2k)!}\sum_{n=0}^{2k}(-1)^n\binom{2k}{n}(k-n)^{2k}=\sum_{n=0}^{2k}\frac{(-1)^n(k-n)^{2k}}{n!(2k-n)!}=2\sum_{n=0}^{k}\frac{(-1)^n(k-n)^{2k}}{n!(2k-n)!}.$$ So it suffices to show that (in our case $j=2k$, and $x=k$) $$\sum_{n=0}^{j}(-1)^n\binom{j}{n}(x-n)^{j}=j!$$ This is a well known identity:

1) Combinatorial proof of $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!$, using inclusion-exclusion

2) Expressing a factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\left.S_{k} \equiv \sum_{n = 0}^{k}{\pars{-1}^{n}\pars{k - n}^{2k} \over n!\pars{2k - n}!}\right\vert_{\ k\ \in\ \mathbb{N}_{\ \geq\ 1}} = {1 \over 2}:\ {\Large ?}}$.

Note that $\ds{S_{k} \equiv \sum_{n = 0}^{\color{red}{k}}{\pars{-1}^{n}\pars{k - n}^{2k} \over n!\pars{2k - n}!} = {1 \over \pars{2k}!}\sum_{n = 0}^{\color{red}{k - 1}} {2k \choose n}\pars{-1}^{n}\pars{k - n}^{2k}\ \mbox{with}\ \color{red}{k} \geq 2}$.

For $\ds{\color{red}{k} = 1}$, the sum $\ds{S_{1}}$ is already equal to $\ds{\large{1 \over 2}}$ !!!.

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$$ \mbox{Note that}\quad\bbx{% \sum_{n = 0}^{k - 1}{\pars{-1}^{n}\pars{k - n}^{2k} \over n!\pars{2k - n}!} = {1 \over 2}\sum_{n = 0}^{2k}{\pars{-1}^{n}\pars{k - n}^{2k} \over n!\pars{2k - n}!}} $$

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\begin{align} &\sum_{n = 0}^{k - 1}{\pars{-1}^{n}\pars{k - n}^{2k} \over n!\pars{2k - n}!} = {1 \over 2\pars{2k}!}\sum_{n = 0}^{2k}{2k \choose n}\pars{-1}^{n} \bracks{\pars{2k}!\oint_{\verts{z}\ =\ 1}{\expo{\pars{n - k}z} \over z^{2k + 1}} \,{\dd z \over 2\pi\ic}} \\[5mm] = &\ {1 \over 2}\oint_{\verts{z}\ =\ 1} {\expo{-kz} \over z^{2k + 1}}\sum_{n = 0}^{2k}{2k \choose n}\pars{-\expo{z}}^{n} \,{\dd z \over 2\pi\ic} = {1 \over 2}\oint_{\verts{z}\ =\ 1} {\expo{-kz} \over z^{2k + 1}}\pars{1 - \expo{z}}^{2k} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ {1 \over 2}\oint_{\verts{z}\ =\ 1} {\expo{-kz} \over z^{2k + 1}}\bracks{\pars{2k}!\sum_{n = 0}^{\infty} {n \brace 2k}{z^{n} \over n!}}\,{\dd z \over 2\pi\ic} = {\pars{2k}! \over 2}\sum_{n = 2k}^{\infty} {n \brace 2k}{1 \over n!}\oint_{\verts{z}\ =\ 1} {\expo{-kz} \over z^{2k - n + 1}}\,{\dd z \over 2\pi\ic} \\[5mm] = &\ {\pars{2k}! \over 2}\sum_{n = 0}^{\infty} {n + 2k \brace 2k}{1 \over \pars{n + 2k}!}{\pars{-kz}^{-n} \over \pars{-n}!} = \left.{\pars{2k}! \over 2} {n + 2k \brace 2k}{1 \over \pars{n + 2k}!}{\pars{-kz}^{-n} \over \pars{-n}!} \right\vert_{\ n\ =\ 0} \\[5mm] = & \bbx{1 \over 2} \end{align}

Answer 3

We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write e.g. \begin{align*} n![z^n]e^{kz}=k^n\tag{1} \end{align*}

We obtain (interchanging $k$ and $n$ for convenience only) \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\frac{(n-k)^{2n}}{k!(2n-k)!}}\\ &=\sum_{k=0}^n(-1)^{n-k}\frac{k^{2n}}{(n-k)!(n+k)!}\tag{2}\\ &=\frac{1}{(2n)!}\sum_{k=0}^n(-1)^{n-k}\binom{2n}{n-k}k^{2n}\tag{3}\\ &=\frac{1}{2(2n)!}\sum_{k=-n}^n(-1)^{n-k}\binom{2n}{n-k}k^{2n}\tag{4}\\ &=\frac{1}{2(2n)!}\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}(k-n)^{2n}\tag{5}\\ &=\frac{1}{2}\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}[z^{2n}]e^{(k-n)z}\tag{6}\\ &=\frac{1}{2}[z^{2n}]e^{-nz}\sum_{k=0}^{2n}\binom{2n}{k}(-e^z)^k\tag{7}\\ &=\frac{1}{2}[z^{2n}]e^{-nz}(1-e^z)^{2n}\tag{8}\\ &=\frac{1}{2}[z^{0}]e^{-nz}\tag{9}\\ &\,\,\color{blue}{=\frac{1}{2}} \end{align*} and the claim follows.

Comment:

• In (2) we exchange the order of summation $k\to n-k$.

• In (3) we use the binomial coefficient notation.

• In (4) we observe that each summand is symmetric with respect to $k$ and we use this property to sum the half of the summands from $-n$ to $n$.

• In (5) we shift the index to start from $k=0$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

• In (6) we apply the coefficient of operator according to (1).

• In (7) we use the linearity of the coefficient of operator and do some arrangements as preparation for the next step.

• In (8) we use the binomial theorem.

• In (9) we note that $[z^{2n}](1-e^z)^{2n}=[z^{2n}](z+\frac{z^2}{2!}+\cdots)^{2n}=1$.