The reciprocal of Pascal Triangle

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Published on 2011. február 01. kedd, 07:57

While all the high school educated people heared about Pascal's triangle, the just as nice Leibnitz-triangle is not very well known.

The rows of Pascal's triangle are conventionally enumerated starting with row $n = 0$ at the top. The entries in each row are numbered from the left beginning with $k = 0$ and are usually staggered relative to the numbers in the adjacent rows. A simple construction of the triangle proceeds in the following manner. On row 0, write only the number 1. Then, to construct the elements of following rows, add the number directly above and to the left with the number directly above and to the right to find the new value. If either the number to the right or left is not present, substitute a zero in its place. This is shown in the picture below:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 . . . . . . . . . . . . . . .

In Pascal's triangle the $k$-th element in the $n$-th row is the binomial coefficient ${n \choose k}$.

Can Pascal's triangle be turned down-side up?

Observe that next to the outer diagonals of the sequence of the natural numbers appears.

The rows of Leibniz's triangle are conventionally enumerated starting with row n = 0 at the bottom. The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. A simple construction of the triangle proceeds in the following manner. On row 0, write only the number 1. Then, to construct the elements of following rows upwards, add the reciprocal of the row's number and proceed so that every number suplements the previous number to the value of the numbers directly down and left to it. This is shown in the picture below:

. . . . . . . . 1/7 1/42 .. .. .. .. 1/7 1/6 1/30 1/60 1/60 1/30 1/6 1/5 1/20 1/30 1/20 1/5 1/4 1/12 1/12 1/4 1/3 1/6 1/3 1/2 1/2 1

Clearly, that each element in Leibniz's triangle is, like in Pascal's triangle, the sum of its two neighbors directly above it. The $k$-th element of Leibniz's triangle in the $n$-th row is given by the reciprocal of $(n+1){n\choose k}$.

The sum of the elements in the diagonal of direction north-east starting from the first element in the $n$-th row is $$\eqalign{ \sum_{j=0}^\infty \frac{1/(j+1+n)}{{j+n\choose j}} &=\lim_{k\to\infty}\sum_{j=0}^k \frac{1}{j+1} \ge \lim_{k\to\infty}\int_{0}^k \frac{1}{x+1} dx = \lim_{k\to\infty}\ln(k+1)=\infty. }$$ if $n=0$. If $n>0$, then a telescopic sum gives the nice result: $$\eqalign{ \sum_{j=0}^\infty \frac{1/(j+1+n)}{{j+n\choose j}} &=n!\lim_{k\to\infty}\sum_{j=0}^k\frac{1}{(j+1)\cdots(j+1+n)} =n!\lim_{k\to\infty}\frac{1}{n}\left(\frac{1}{n!}-\frac{(k+1)!}{(k+n+1)!}\right) \\ &=\lim_{k\to\infty}\frac{1}{n}\left(1-\frac{1}{{k+n+1\choose n}}\right) =\frac{1}{n}. }$$

Some have indicated that it would be good to point some details of the derivation above....

So, here are some interesting formula that might help or not. :-)

Through a simple calculation we get $$ \frac{n!}{j(j+1)\cdots(j+n)} ={n\choose 0}\frac{1}{j}+\cdots+(-1)^{\ell}{n\choose\ell}\frac{1}{j+\ell}+\cdots+(-1)^{n}{n\choose n}\frac{1}{j+n}\ . $$ This yields $$ \frac{n!}{j(j+1)\cdots(j+n)} =\int_0^1x^{j-1}(1-x)^n\,dx, $$ so we have $$\eqalign{ \sum_{j=0}^\infty \frac{1/(j+1+n)}{{j+n\choose j}} &=\lim_{k\to\infty}\sum_{j=0}^k\frac{n!}{(j+1)\cdots(j+1+n)} =\lim_{k\to\infty}\int_0^1(1-x^{k})(1-x)^{n-1}\,dx \\ &=\int_0^1(1-x)^{n-1}\,dx =\frac{1}{n}. }$$