Kishlaya's Identity

In this paper, I'll be presenting and proving a new identity which I dub as "\(\textbf{Kishlaya's Identity}\)". I happened upon it while investigating on the Binomial Coefficients in the denominator. It proves great in finding infinite sums where binomial coefficients appear in the denominator. Also, a special thanks to Cody Johnson for helping me out to simplify the identity and coming up with another interesting result from this identity. Also, shoutout to a Brilliant user, Jon Haussmann, who came up with an another amazing proof for this identity.

For all \(|x|<1\)

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m}{n+m \choose m}^{-1} = \log\left(\frac{1}{1-x}\right)\]

Convergence Test

Ratio Test

We check if the following limit exists and results in a finite value \[\lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \frac{a_{n,m}}{a_{n-1,m-1}}\] where \[a_{n,m} = \frac{x^n}{m{n+m \choose m}} = \frac{n!\ (m-1)!}{(n+m)!}x^n\] \[\begin{eqnarray} \therefore \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \left(\frac{n!\ (m-1)!}{(n+m)!}x^n\right)\left(\frac{(n+m-2)!}{(n-1)!\ (m-2)!}\frac{1}{x^{n-1}}\right) & = & \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \frac{xn(m-1)}{(n+m)(n+m-1)} \nonumber \\ & = & \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \frac{nx}{m} - \frac{2n^2x}{m^2} + \frac{n^2(3n-1)x}{m^3} + O\left(\frac{1}{m^4}\right) \nonumber \\ & = & 0 \nonumber \end{eqnarray}\]

Thus, we can conclude via the ratio test that the series converges.

We observe the denominator closely. \[\frac{1}{m{n+m \choose m}} = \frac{n!\ (m-1)!}{(n+m)!} = B(m,n+1) = \int_0^1 t^n(1-t)^{m-1}\ dt\] where \(B(x,y)\) denotes the Beta function. Thus, we can express the sum as \[\begin{eqnarray} \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} & = & \sum_{n=1}^\infty \sum_{m=1}^\infty \int_0^1 (xt)^n(1-t)^{m-1}\ dt \nonumber \\ & = & \int_0^1 \left(\sum_{n=1}^\infty (xt)^n\right) \left(\sum_{m=1}^\infty (1-t)^{m-1}\right)\ dt \nonumber \end{eqnarray}\] Now, since \(x\ \in (-1,1)\), we can use the Geometric Expansion that is \(\sum_{n=1}^\infty r^n = \frac{r}{1-r}\), to condense the sum as \[\begin{eqnarray} \int_0^1 \left(\sum_{n=1}^\infty (xt)^n\right) \left(\sum_{m=1}^\infty (1-t)^{m-1}\right)\ dt & = & \int_0^1 \left(\frac{xt}{1-xt}\right) \left(\frac{1}{t}\right)\ dt \nonumber \\ & = & \int_0^1 \frac{x}{1-xt}\ dt \nonumber \end{eqnarray}\] The only work left is to evaluate the integral \[\begin{eqnarray} \int_0^1 \frac{x}{1-xt}\ dt & = & -\log(1-xt)\vert_0^1 \nonumber \\ & = & \log\left(\frac{1}{1-x}\right) \nonumber \end{eqnarray}\] And thus, \[\begin{equation} \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} = \log\left(\frac{1}{1-x}\right) \end{equation}\]

\(\textbf{Proof By Jon Haussmann}\) \[\begin{eqnarray} \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} & = & \sum_{n=1}^\infty x^n \sum_{m=1}^\infty \frac{(m-1)!\ n!}{(m+n)!} \nonumber \\ & = & \sum_{n=1}^\infty \frac{x^n}{n} \sum_{m=1}^\infty \frac{(m-1)!\ n!\ n}{(m+n)!} \nonumber \end{eqnarray}\] Now, we try telescoping the following series : \[\begin{eqnarray} \sum_{m=1}^\infty \frac{(m-1)!\ n!\ n}{(m+n)!} & = & \sum_{m=1}^\infty \frac{(m-1)!\ n!\ ((m+n)-m)}{(m+n)!} \nonumber \\ & = & \sum_{m=1}^\infty \frac{(m-1)!\ n!\ (m+n)}{(m+n)!} - \sum_{m=1}^\infty \frac{(m-1)!\ n!\ m}{(m+n)!} \nonumber \\ & = & \sum_{m=1}^\infty \frac{(m-1)!\ n!}{(m+n-1)!} - \sum_{m=1}^\infty \frac{m!\ n!}{(m+n)!} \nonumber \\ & = & n!\left(\sum_{m=1}^\infty \frac{(m-1)!}{(m+n-1)!} - \sum_{m=1}^\infty \frac{m!}{(m+n)!}\right) \nonumber \\ & = & n!\left(\frac{0!}{n!}\right) \nonumber \\ & = & 1 \nonumber \\ \therefore \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} & = & \sum_{n=1}^\infty \frac{x^n}{n} \nonumber \end{eqnarray}\] And thus, \[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} = \log\left(\frac{1}{1-x}\right)\]

The special case when \(n=m\), the double sum condenses to a single one and we get the following result \[\sum_{n=1}^\infty \frac{x^n}{n}{2n \choose n}^{-1} = 2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)\]

We proceed in a similar manner as we did earlier.

We write the denominator as \[\frac{1}{n{2n \choose n}} = \frac{n!\ (n-1)!}{(2n)!} = B(n,n+1) = \int_0^1 t^n(1-t)^{n-1}\ dt\] Now, we can express the sum as \[\begin{eqnarray} \sum_{n=1}^\infty \frac{x^n}{n{2n \choose n}} & = & \sum_{n=1}^\infty \int_0^1 \left(xt(1-t)\right)^n\frac{1}{1-t}\ dt \nonumber \\ & = & \int_0^1 \left(\sum_{n=1}^\infty \left(xt(1-t)\right)^n\right)\frac{1}{1-t}\ dt \nonumber \\ & = & \int_0^1 \frac{xt}{1+xt(1-t)}\ dt \nonumber \end{eqnarray}\] The only work left is to evaluate the integral.

Let, \[\begin{eqnarray} I & = & \int_0^1 \frac{xt}{1+xt(1-t)}\ dt \nonumber \\ & = & \int_0^1 \frac{x(1-t)}{1+xt(1-t)}\ dt \nonumber \\ & = & \int_0^1 \frac{x}{1+xt(1-t)}\ dt - \int_0^1 \frac{xt}{1+xt(1-t)}\ dt \nonumber \\ & = & \int_0^1 \frac{x}{1+xt(1-t)}\ dt - I \nonumber \\ \therefore\ I & = & \frac{x}{2}\int_0^1 \frac{dt}{1+xt(1-t)} \nonumber \\ & = & \frac{1}{2}\int_0^1 \frac{dt}{\left(t-\frac{1}{2}\right)^2 + \left(\sqrt{\frac{1}{x}-\frac{1}{4}}\right)^2} \nonumber \\ & = & 2\sqrt{\frac{x}{4-x}}\tan^{-1}\left(\frac{x}{4-x}\right) \nonumber \\ & = & 2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}{2}\right) \nonumber \end{eqnarray}\] And thus, \[\begin{equation} \sum_{n=1}^\infty \frac{x^n}{n{2n \choose n}} = 2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}{2}\right) \end{equation}\]

And therefore, the result follows. \(\square\)

Kishlaya's Identity can be generalized for an even number of variables as follows

\[\sum_{a_{2n}, a_{2n-1},\ldots, a_1 \geq 1} \frac{x^{a_{2n}}}{(a_1+a_2+\ldots+a_n)}{a_1+a_2+\ldots+a_{2n} \choose a_1+a_2+\ldots+a_n}^{-1} = \log\left(\frac{1}{1-x}\right)\]

For odd number of variables, the result follows as

\[\sum_{a_{2n+1}, a_{2n},\ldots, a_1 \geq 1} \frac{x^{a_{2n+1}}}{(a_{n+1}+a_{n+2}+\ldots+a_{2n+1})}{a_1+a_2+\ldots+a_{2n+1} \choose a_{1}+a_{2}+\ldots+a_{n}}^{-1} = \log\left(\frac{1}{1-x}\right)\]

Evaluate

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{n!\ (m-1)!}{2^n\ (n+m)!}\]

\(\textbf{Solution.}\) Setting \(x=\frac{1}{2}\) in the Kishlaya's Identity yields

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{2^n\ m{n+m \choose m}} = \log\left(\frac{1}{1-\frac{1}{2}}\right)\]

\[\Rightarrow\ \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{n!\ (m-1)!}{2^n\ (n+m)!} = \log(2)\]

Prove for all \(|x| < 1\)

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nx^n}{m{n+m \choose m}} = \sum_{k=1}^\infty x^k\]

\(\textbf{Solution.}\) We first try to simplify the sum on the R.H.S. Observe that \[\sum_{k=1}^\infty x^k = x+x^2+x^3+\ldots = x(1+x+x^2+\ldots) = \frac{x}{1-x}\]

After canceling out the common factor (i.e. \(x\)) on both sides, it suffices to prove that

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nx^{n-1}}{m{n+m \choose m}} = \frac{1}{1-x}\]

Now, we observe that the \(nx^{n-1}\) in the numerator on L.H.S is the derivative of \(x^n\) which gives us a clue to differentiate the Kishlaya's Identity w.r.t \(x\) and we get

\[\frac{d}{dx}\left(\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}}\right) = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{\frac{d(x^n)}{dx}}{m{n+m \choose m}} = \frac{d}{dx}(-\log(1-x))\]

\[\Rightarrow \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nx^{n-1}}{m{n+m \choose m}} = \frac{1}{1-x}\]

Hence, proved.

Prove Lehmer's Identity, for all \(|x| < 1\)

\[\sum_{n=1}^\infty \frac{(2x)^{2n}}{n{2n \choose n}} = \frac{2x}{\sqrt{1-x^2}}\sin^{-1}(x)\]

\(\textbf{Solution.}\) We make the substitution \(x \rightarrow 4x^2\) in special case of Kishlaya's Identity and the result follows.

Cody Johnson, came up with an interesting observation related to Kishlaya's Identity.

\[\sum_{m=1}^\infty \frac{1}{m{n+m \choose m}} = \frac{1}{n}\]

We again call Beta Function for help.

\[\begin{eqnarray} \sum_{m=1}^\infty \frac{1}{m{n+m \choose m}} & = & \sum_{m=1}^\infty \frac{n!\ (m-1)!}{(n+m)!} \nonumber \\ & = & \sum_{m=1}^\infty \int_0^1 t^n(1-t)^{m-1} \nonumber \\ & = & \int_0^1 t^{n-1} \nonumber \\ & = & \frac{1}{n} \nonumber \\ \end{eqnarray}\] And thus, once again we thank the Beta Function.

\(\textbf{Proof By Cody Johnson}\)

We make use of the MacLaurin Expansion of \(\log(1-x)\) \[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} = -\log(1-x)\] \[\Rightarrow \sum_{n=1}^\infty x^n \left(\sum_{m=1}^\infty \frac{1}{m{n+m \choose m}}\right) = \sum_{k=1}^\infty \frac{x^k}{k}\] On comparing the coefficients, we can conclude the following \[\sum_{m=1}^\infty \frac{1}{m{n+m \choose m}} = \frac{1}{n}\]

\(\textbf{Problem 1.}\) Prove \[\sum_{n,m \geq 1} \frac{x^n}{{n+m \choose m-1}} = \frac{x}{1-x}-\log(1-x)\]

\(\textbf{Problem 2.}\) Prove \[\sum_{n,m \geq 1} \frac{x^n}{(n+m+1){n+m \choose m}} = 1+\left(\frac{1-x}{x}\right)\log(1-x)\]