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Show that given a double series $a_{m,n}=\frac{m-n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}$, where $a_{0,0}$ is defined to be zero, its sum by rows $\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}$ is -1, its sum by columns $\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{m,n}$ is 1.

My thoughts are as follows:

  1. Let $S_{M,N}=\sum_{m=0, n=0}^{M, N}a_{m,n}$, then it's obvious that $a_{m,n}=-a_{n,m}$, $S_{M,N}=-S_{N,M}$, and $\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n} =- \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{m,n}$.

  2. I guessed $S_{M,N}=\frac{M-N}{M+N}$, it seems to be supported by the observation that $a_{m,n}=\frac{1}{2^{m+n}}\frac{m-n}{m+n}\frac{(m+n)!}{m!n!}$, (where ${n+m\choose m}$ and $2^{m+n}$ might cancel out in summation), but if so, $a_{m,n}$ should be $(\frac{1}{m+n}+\frac{2}{m+n-1}+\frac{1}{m+n-2})(m-n)$, which contradicts the above definition of $a_{m,n}$.

With k=m+n, l=m-n (l=-k,-k+2,..., k), we can further write the series as $$a_{m,n}=a'_{k,l}=\frac{1}{2^{k}}\frac{l}{k}{k\choose\frac{k+l}{2}},$$ the part without l/k is exactly the expansion of $(1+1)^k/2^k$ for each k.

  1. Then I tried to use induction as well as recursion relations $a_{m+1,n}=\frac{(m+1-n) (m+n) }{(m-n) (m+1) (2) }a_{m,n}$ and $a_{m,n+1}=\frac{(m-n-1) (m+n)}{(m-n) (n+1) (2) }a_{m,n}$ to get sum $R_n$ of items in (n+1)th row.

$R_0=\sum_{n=0}^{\infty}a_{0,n}=a_{0,0}+\sum_{n=1}^{\infty}-2^{-n}=-1,$

$R_1=\sum_{n=0}^{\infty}a_{1,n}=\sum_{n=0}^{\infty}\frac{n-1}{2} a_{0,n}=\sum_{n=0}^{\infty}\frac{n-1}{2} (-2^{-n})=0,$ using relation between $a_{m,n}$ and $a_{m+1,n}$.

I tried to simplify $R_2$ to make it a funcion of $a_{1,n}$ or $a_{0,n}$ and $n$, using relationship between $a_{m,n}$ and $a_{m+1,n}$ as well as that between $a_{m,n}$ and $a_{m,n+1}$, but the result I get can't yet make $R_2$ a function of $R_1$ or $R_0$. I guess it should be 0 but I haven't proved that yet.

Charlie Chang
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2 Answers2

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For $m>0$, your $R_m$ can be computed using the binomial series: for $k\geqslant 0$ and $|z|<1$ $$\sum_{n=0}^\infty\frac{(n+k)!}{n!\,k!}z^k=(1-z)^{-k-1}\implies\sum_{n=0}^\infty\frac{(n+k)!}{2^n\,n!\,k!}=2^{k+1},$$ giving \begin{align*} R_m&=\frac{1}{2^m}\left(\sum_{n=0}^\infty\frac{(m+n-1)!}{2^n\,(m-1)!\,n!}-\sum_{n=1}^\infty\frac{(m+n-1)!}{2^n\,m!\,(n-1)!}\right) \\&=\frac{1}{2^m}\left(\sum_{n=0}^\infty\frac{(n+m-1)!}{2^n\,n!\,(m-1)!}-\sum_{n=0}^\infty\frac{(n+m)!}{2^{n+1}\,n!\,m!}\right) \\&=\frac{1}{2^m}\left(2^m-\frac12\cdot 2^{m+1}\right)=\color{red}{0}. \end{align*} All the rest is already done by you.

metamorphy
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This solution is inspired by the answer @metamorphy.

Notice $a_{m+1,n}+a_{m,n+1}=\frac{(m+1-n) (m+n) }{(m-n) (m+1) (2) }+\frac{(m-n-1) (m+n)}{(m-n) (n+1) (2) }a_{m,n} =\frac{(m+n+1) (m+n)}{(m+1) (n+1) (2) }a_{m,n}=2a_{m+1,n+1},$

so we have $(R_0-a_{0,0})+R_1=2\sum_{n=1}^{\infty}a_{1,n}=2(R_1-a_{1,0})$, and so $R_1=R_0-a_{0,0}+2a_{1,0}=-1-0+2\cdot1/2=0$.

Similarly, $R_1-a_{1,0})+R_2=2(R_2-a_{2,0})$, which leads to $R_2=R_1-a_{1,0}+2a_{2,0}=0$, for $a_{1,0}=2a_{2,0}$. Thus we get $R_3=R_4=\dots=0$, and so $\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n}=\sum_{m=0}^{\infty}R_m=0$.

The sum by columns should be calculated almost the same way; or we can use the mentioned 'anti-symmetry', i.e. either sum of $n$th column equals negative of sum of $n$th row, or, sum by columns equals negative of sum by rows.

So we see indeed there is a clear recursive relation between sums of adjacent rows/columns, as a result of combinatorics. This together with 'anti-symmetry', caused by 'm-n', gives the solution. Besides, notice, a bit unnaturally, $a_{0,0}$(beside all diagonal elements) has to be zero as a result of 'anti-symmetry'.

I guess if I had written out adjacent elements in a few rows then the recursive relation could have been quite obvious.

From my above recursive relations, it seems all elements are powers of 1/2, which seems not true, e.g. $a_{3,5}$. After double check, I see not all elements but those near the diagonal line and those in first column/row are powers of 1/2, e.g. $a_{1,2}=a_{1,3}=-1/8$ while $a_{1,4}=\frac{-1/8-1/16}{2}=\frac{-3}{32}$. The further an element is from the diagonal line and from edges, the more 'irregular' it becomes. This property is also quite like the binomial triangle, so I guess this double series (considering its upper or lower half triangle) are somehow like extension of/distorted binomial triangle, and one of a class of (algebraic) structure.

Charlie Chang
  • 1,575