16. Let the function $f(x)=p x-\frac{p}{x}-2 \ln x$.
(1) If $f(x)$ is a monotonically increasing function in its domain, find the range of the real number $p$;
(2) Let $g(x)=\frac{2 \mathrm{e}}{x}$, and $p>0$. If there exists at least one point $x_{0}$ in the interval $[1, \mathrm{e}]$ such that $f\left(x_{0}\right)>g\left(x_{0}\right)$, find the range of the real number $p$;
(3) Prove that for any positive integer $n$, 
$$
\sum_{k=1}^{n} \ln ^{2}\left(1+\frac{2}{k}\right)<3 \text {. }
$$
16. (1) The domain of the function $f(x)$ is $\{x \mid x>0\}$. To make $f(x)$ a monotonically increasing function in its domain $(0,+\infty)$, it is sufficient that
$$
f^{\prime}(x)=p+\frac{p}{x^{2}}-\frac{2}{x} \geqslant 0
$$

holds for all $x$ in the interval $(0,+\infty)$, i.e.,
$$
p \geqslant\left(\frac{2 x}{x^{2}+1}\right)_{\max }=1 \text {. }
$$
(2) The original proposition is equivalent to $f(x)-g(x)>0$ having a solution in the interval $[1, \mathrm{e}]$.
$$
\begin{array}{l}
\text { Let } F(x)=f(x)-g(x) \\
=p x-\frac{p}{x}-2 \ln x-\frac{2 \mathrm{e}}{x} . \\
\text { By } F^{\prime}(x)=p+\frac{p}{x^{2}}-\frac{2}{x}+\frac{2 \mathrm{e}}{x^{2}} \\
=\frac{p x^{2}+p+2(\mathrm{e}-x)}{x^{2}}>0
\end{array}
$$

From $F^{\prime}(x)=p+\frac{p}{x^{2}}-\frac{2}{x}+\frac{2 \mathrm{e}}{x^{2}}$
$\Rightarrow F(x)$ is an increasing function
$$
\begin{array}{l}
\Rightarrow F(x)_{\max }=F(\mathrm{e})>0 \\
\Rightarrow p>\frac{4 \mathrm{e}}{\mathrm{e}^{2}-1} \\
\Rightarrow p \in\left(\frac{4 \mathrm{e}}{\mathrm{e}^{2}-1},+\infty\right) .
\end{array}
$$
(3) Let $g(x)=2 \ln x-x+\frac{1}{x}$.

Then, from (1), we know that $g(x)$ is a monotonically decreasing function in the interval $(0,+\infty)$.
By $g(1)=0$, we know that when $x>1$, $g(x)<0$, i.e.,
$$
0<2 \ln x<x-\frac{1}{x}=\frac{x^{2}-1}{x} \text {. }
$$

Thus, $2 \ln \sqrt{1+\frac{2}{k}}<\frac{2}{\sqrt{k(k+2)}}$, i.e.,
$$
\begin{array}{l}
\ln \left(1+\frac{2}{k}\right)<\frac{2}{\sqrt{k(k+2)}} . \\
\text { Therefore, } \sum_{k=1}^{n} \ln ^{2}\left(1+\frac{2}{k}\right)<\sum_{k=1}^{n} \frac{4}{k(k+2)} \\
=2 \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+2}\right) \\
=2\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)<3 .
\end{array}
$$