= Solution
Taking reciprocals in the finite product defining the <gamma function> gives
$$
\frac1{\Gamma(z)}
=z\lim_{n\to\infty}n^{-z}
\prod_{k=1}^n\left(1+\frac zk\right).
$$
Let $H_n=\sum_{k=1}^n k^{-1}$. Since
$$
n^{-z}
=e^{z(H_n-\log n)}e^{-zH_n}
=e^{z(H_n-\log n)}\prod_{k=1}^n e^{-z/k}
$$
and $H_n-\log n\to\gamma$, where $\gamma$ is the <Euler--Mascheroni constant>, passage to the limit gives the <Weierstrass product for the reciprocal gamma function>
$$
\boxed{
\frac1{\Gamma(z)}
=ze^{\gamma z}\prod_{k=1}^{\infty}
\left(1+\frac zk\right)e^{-z/k}
}.
$$
The logarithmic derivative of this identity is
$$
-\psi(z)
=\frac1z+\gamma
+\sum_{k=1}^{\infty}
\left(\frac1{z+k}-\frac1k\right).
$$
Since $k^{-1}-(z+k)^{-1}=z/[k(z+k)]$, the <digamma function> therefore satisfies
$$
\boxed{
\psi(z)=-\gamma-\frac1z
+z\sum_{k=1}^{\infty}\frac1{k(z+k)}
}.
$$
For real $z>0$, termwise differentiation gives the <trigamma function>
$$
\psi'(z)=\frac1{z^2}
+\sum_{k=1}^{\infty}\frac1{(z+k)^2}>0.
$$
Thus $\psi$ is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping gives
$$
\psi(1)=-\gamma<0,
\qquad
\psi(2)=1-\gamma>0.
$$
The <intermediate value theorem> supplies a zero in $(1,2)$, and strict increase makes it unique. This is the <positive zero of the digamma function>.
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