Egorychev method

Suppose we seek to evaluate

${\displaystyle S_{j}(n)=\sum _{k=0}^{n}(-1)^{k}{n \choose k}{n+k \choose k}{k \choose j}}$

which is claimed to be :${\displaystyle (-1)^{n}{n \choose j}{n+j \choose j}.}$

Introduce :${\displaystyle {n+k \choose k}={\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n+k}}{z^{k+1}}}\;dz}$

and :${\displaystyle {k \choose j}={\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {(1+w)^{k}}{w^{j+1}}}\;dw.}$

This yields for the sum :

${\displaystyle {\begin{aligned}&{\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n}}{z}}{\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{j+1}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(1+z)^{k}(1+w)^{k}}{z^{k}}}\;dw\;dz\\[6pt]={}&{\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n}}{z}}{\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{j+1}}}\left(1-{\frac {(1+w)(1+z)}{z}}\right)^{n}\;dw\;dz\\[6pt]={}&{\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n}}{z^{n+1}}}{\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{j+1}}}(-1-w-wz)^{n}\;dw\;dz\\[6pt]={}&{\frac {(-1)^{n}}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n}}{z^{n+1}}}{\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{j+1}}}(1+w+wz)^{n}\;dw\;dz.\end{aligned}}}$

This is

${\displaystyle {\frac {(-1)^{n}}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n}}{z^{n+1}}}{\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{j+1}}}\sum _{q=0}^{n}{n \choose q}w^{q}(1+z)^{q}\;dw\;dz.}$

Extracting the residue at ${\displaystyle w=0}$ we get

${\displaystyle {\begin{aligned}&{\frac {(-1)^{n}}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n}}{z^{n+1}}}{n \choose j}(1+z)^{j}\;dz\\[6pt]={}&{n \choose j}{\frac {(-1)^{n}}{2\pi i}}\int _{|z|=\varepsilon }{\frac {(1+z)^{n+j}}{z^{n+1}}}\;dz\\[6pt]={}&(-1)^{n}{n \choose j}{n+j \choose n}\end{aligned}}}$

thus proving the claim.

Suppose we seek to evaluate ${\displaystyle \sum _{k=1}^{n}k{2n \choose n+k}.}$

Introduce

${\displaystyle {2n \choose n+k}={\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {1}{z^{n-k+1}}}{\frac {1}{(1-z)^{n+k+1}}}\;dz.}$

Observe that this is zero when ${\displaystyle k>n}$ so we may extend ${\displaystyle k}$ to infinity to obtain for the sum

${\displaystyle {\begin{aligned}&{\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {1}{z^{n+1}}}{\frac {1}{(1-z)^{n+1}}}\sum _{k\geq 1}k{\frac {z^{k}}{(1-z)^{k}}}\;dz\\[6pt]={}&{\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {1}{z^{n+1}}}{\frac {1}{(1-z)^{n+1}}}{\frac {z/(1-z)}{(1-z/(1-z))^{2}}}\;dz\\[6pt]={}&{\frac {1}{2\pi i}}\int _{|z|=\varepsilon }{\frac {1}{z^{n}}}{\frac {1}{(1-z)^{n}}}{\frac {1}{(1-2z)^{2}}}\;dz.\end{aligned}}}$

Now put ${\displaystyle z(1-z)=w}$ so that (observe that with ${\displaystyle w=z+\cdots }$ the image of ${\displaystyle |z|=\varepsilon }$ with ${\displaystyle \varepsilon }$ small is another closed circle-like contour which we may certainly deform to obtain another circle ${\displaystyle |w|=\gamma }$)

${\displaystyle z={\frac {1-{\sqrt {1-4w}}}{2}}\quad {\text{and}}\quad (1-2z)^{2}=1-4w}$

and furthermore

${\displaystyle dz=-{\frac {1}{2}}\times {\frac {1}{2}}\times (-4)\times (1-4w)^{-1/2}\;dw=(1-4w)^{-1/2}\;dw}$

to get for the integral

${\displaystyle {\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{n}}}{\frac {1}{1-4w}}(1-4w)^{-1/2}\;dw={\frac {1}{2\pi i}}\int _{|w|=\gamma }{\frac {1}{w^{n}}}{\frac {1}{(1-4w)^{3/2}}}\;dw.}$

This evaluates by inspection to (use the Newton binomial)

${\displaystyle {\begin{aligned}&4^{n-1}{n-1+1/2 \choose n-1}=4^{n-1}{n-1/2 \choose n-1}={\frac {4^{n-1}}{(n-1)!}}\prod _{q=0}^{n-2}(n-1/2-q)\\={}&{\frac {2^{n-1}}{(n-1)!}}\prod _{q=0}^{n-2}(2n-2q-1)={\frac {2^{n-1}}{(n-1)!}}{\frac {(2n-1)!}{2^{n-1}(n-1)!}}\\[6pt]={}&{\frac {n^{2}}{2n}}{2n \choose n}={\frac {1}{2}}n{2n \choose n}.\end{aligned}}}$

Here the mapping from ${\displaystyle z=0}$ to ${\displaystyle w=0}$ determines the choice of square root. This example also yields to simpler methods but was included here to demonstrate the effect of substituting into the variable of integration.

• Computational examples of using the Egorychev method to evaluate sums involving binomial coefficients

• Egorychev, G. P. (1984). Integral representation and the Computation of Combinatorial sums. American Mathematical Society.