Solution

Taking reciprocals in the finite product defining the gamma function gives

\frac{1}{Γ ( z )} = z lim_{n \to \infty} n^{- z} \prod_{k = 1}^{n} (1 + \frac{z}{k}) .

Let

H_{n} = \sum_{k = 1}^{n} k^{- 1}

. Since

n^{- z} = e^{z (H_{n} - l o g n)} e^{- z H_{n}} = e^{z (H_{n} - l o g n)} \prod_{k = 1}^{n} e^{- z / k}

and

H_{n} - lo g n \to γ

, where

γ

\frac{1}{Γ ( z )} = z e^{γ z} k = 1 \prod \infty (1 + \frac{z}{k}) e^{- z / k} .

The logarithmic derivative of this identity is

- ψ (z) = \frac{1}{z} + γ + \sum_{k = 1}^{\infty} (\frac{1}{z + k} - \frac{1}{k}) .

Since

k^{- 1} - (z + k)^{- 1} = z / [k (z + k)]

, the digamma function therefore satisfies

ψ (z) = - γ - \frac{1}{z} + z k = 1 \sum \infty \frac{1}{k ( z + k )} .

For real

z > 0

, termwise differentiation gives the trigamma function

ψ^{'} (z) = \frac{1}{z ^{2}} + \sum_{k = 1}^{\infty} \frac{1}{( z + k ) ^{2}} > 0.

Thus

ψ

is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping gives

ψ (1) = - γ < 0, ψ (2) = 1 - γ > 0.

(1, 2)

, and strict increase makes it unique. This is the positive zero of the digamma function.

Solved by gpt-5.6-sol high.

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