A CURIOUS IDENTITY PROVED BY CAUCHY’S INTEGRAL FORMULA

HELMUT PRODINGER^{†}

Simons [2] has proved the identity

∑_{r=0}^{q} \frac{(−1)^{q+r}(q + r)!(1 + x)^{r}}{(q − r)!r!^{2}} = ∑_{r=0}^{q} \frac{(q + r)!x^{r}}{(q − r)!r!^{2}} ;  (1)

Chapman [1] gave a nice and short proof of it. In this note, I want to give another attractive proof. It uses Cauchy’s integral formula to pull out coefficients of generating functions.

We divide (1) by q! and prove the equivalent version

S = ∑_{r=0}^{q} \binom{q}{r}\binom{q + r}{r}(−1)^{q+r}(1 + x)^{r} = ∑_{r=0}^{q} \binom{q}{r}\binom{q + r}{r}x^{r}.

We start with the righthand-side:

S = [t^{q}] ∑_{i≥0} \binom{q}{i}t^{i} · ∑_{i≥0} \binom{q + i}{i}(tx)^{i}

= [t^{q}](1 + t)^{q} · (1 − tx)^{−q−1}

= \frac{1}{2πi} ∮ \frac{dt}{t^{q+1}} (1 + t)^{q} · (1 − tx)^{−q−1}.

Now we substitute t = u/(1 − u), so that dt = du/(1 − u)^{2} and obtain

S = \frac{1}{2πi} ∮ \frac{du}{(1 − u)^{2}} \frac{(1 − u)^{q+1}}{u^{q+1}} (1 − u)^{−q} · (\frac{1 − u(1 + x)}{1 − u})^{−q−1}

= [u^{q}](1 − u)^{q}(1 − u(1 + x))^{−q−1}

= ∑_{r=0}^{q} \binom{−q − 1}{r}(−1)^{r}(1 + x)^{r}\binom{q}{q − r}(−1)^{q−r}

= ∑_{r=0}^{q} \binom{q + r}{r}\binom{q}{r}(1 + x)^{r}(−1)^{q−r},

which is the lefthand-side.

Date: May 8, 2003.

†Supported by NRF Grant 2053748.

References

[1] R. Chapman. A curious identity revisited. The Mathematical Gazette, 87:139–141, 2003.

[2] S. Simons. A curious identity. The Mathematical Gazette, 85:296–298, 2001.

H. Prodinger, The John Knopfmacher Centre for Applicable Analysis and Number Theory, School of Mathematics, University of the Witwatersrand, P. O. Wits, 2050 Johannesburg, South Africa

  E-mail address: helmut@maths.wits.ac.za