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Ira Gessel
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The formula $$\sum_{n=0}^\infty \frac{1}{kn+1}\binom{(k+1)n}{n}(x(1-x)^k)^n=\frac{1}{1-x},\tag{1}$$$$\sum_{n=0}^\infty \frac{1}{kn+1}\binom{(k+1)n}{n}(x(1-x)^k)^n=\frac{1}{1-x}\tag{1}$$ is well known and an easy consequence of Lagrange inversion. If $k$ is a positive integer $(1)$ may be written as $$_kF_{k-1}\left[\frac{1}{k+1},\dots,\frac{k}{k+1};\frac{2}{k},\frac{3}{k},\dots,\frac{k-1}{k},\frac{k+1}{k};\frac{(k+1)^{k+1}}{k^k} x(1-x)^k\right]=\frac{1}{1-x}.$$

Replacing $x$ with $x^k$ gives Glasser's formula.

The formula $$\sum_{n=0}^\infty \frac{1}{kn+1}\binom{(k+1)n}{n}(x(1-x)^k)^n=\frac{1}{1-x},\tag{1}$$ is well known and an easy consequence of Lagrange inversion. If $k$ is a positive integer $(1)$ may be written as $$_kF_{k-1}\left[\frac{1}{k+1},\dots,\frac{k}{k+1};\frac{2}{k},\frac{3}{k},\dots,\frac{k-1}{k},\frac{k+1}{k};\frac{(k+1)^{k+1}}{k^k} x(1-x)^k\right]=\frac{1}{1-x}.$$

Replacing $x$ with $x^k$ gives Glasser's formula.

The formula $$\sum_{n=0}^\infty \frac{1}{kn+1}\binom{(k+1)n}{n}(x(1-x)^k)^n=\frac{1}{1-x}\tag{1}$$ is well known and an easy consequence of Lagrange inversion. If $k$ is a positive integer $(1)$ may be written as $$_kF_{k-1}\left[\frac{1}{k+1},\dots,\frac{k}{k+1};\frac{2}{k},\frac{3}{k},\dots,\frac{k-1}{k},\frac{k+1}{k};\frac{(k+1)^{k+1}}{k^k} x(1-x)^k\right]=\frac{1}{1-x}.$$

Replacing $x$ with $x^k$ gives Glasser's formula.

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Ira Gessel
  • 17k
  • 1
  • 58
  • 80

The formula $$\sum_{n=0}^\infty \frac{1}{kn+1}\binom{(k+1)n}{n}(x(1-x)^k)^n=\frac{1}{1-x},\tag{1}$$ is well known and an easy consequence of Lagrange inversion. If $k$ is a positive integer $(1)$ may be written as $$_kF_{k-1}\left[\frac{1}{k+1},\dots,\frac{k}{k+1};\frac{2}{k},\frac{3}{k},\dots,\frac{k-1}{k},\frac{k+1}{k};\frac{(k+1)^{k+1}}{k^k} x(1-x)^k\right]=\frac{1}{1-x}.$$

Replacing $x$ with $x^k$ gives Glasser's formula.