Why does $\lim_{k \to \infty}\psi(k)$ diverge?

Question

This question is motivated by the fact that $$\psi(k)+\gamma=H_{k-1}$$ This, of course, implies that $$\lim_{k \to \infty}\psi(k)+\gamma = \lim_{k \to \infty}H_{k-1}$$ Since $H_k$ diverges, it follows that $\psi(k+1)$ should also diverge. But $\psi(k)=\frac{\Gamma'(k)}{\Gamma(k)}$. It is intuitive that $\Gamma'(k)<\Gamma(k)$ since the derivative of a function shouldn't grow faster than the function itself, so the denominator would grow faster than the numerator, implying that the limit does not diverge. What is the particular reason that the digamma function diverges?

Answer 1

$$\psi(n)=H_{n-1}+\gamma=\sum_{k=1}^{n-1} \frac{1}{k}+\gamma$$ Notice that $\ln(1+x) <x, x>0$, let $x=1/k$, so $$\frac{1}{k} > \ln(k+1)- \ln k$$ Now we can do telescopic summing Then $$H_{n-1}=\sum_{k=1}^{n-1} \frac{1}{k}>\ln n$$ So when $n\rightarrow \infty$ $H_{n-1}$ and hence $\psi(n)$ diverge.

Answer 2

What's convincing here depends on your precise definitions and what detailed facts about the gamma function you are comfortable with. Perhaps this argument will suit you.

If $\psi(x)=\frac{d}{dx}\log\Gamma(x)$ were bounded above by $C$ for all $x$ sufficiently large, we would have, for some $A>0$ that $\log\Gamma(x)< A+Cx$ for all $x$ sufficiently large, that is, $\Gamma(x)<\exp(A+Cx)$ for all sufficiently large $x$. But the well known identity $\Gamma(x+1)=x\Gamma(x)$ implies that $\Gamma$ grows faster than this.