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 $g(x)=\frac{2 e}{x}$, and $p>0$, if there exists at least one point $x_{0}$ in $[1, e]$ 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 $n$, $\sum_{k=1}^{n} \ln ^{2}\left(1+\frac{2}{k}\right)<3$ holds.
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 $g(x)=\frac{2 e}{x}$, and $p>0$, if there exists at least one point $x_{0}$ in $[1, e]$ 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 $n$, $\sum_{k=1}^{n} \ln ^{2}\left(1+\frac{2}{k}\right) < 3$.
{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 $(0,+\infty)$.

Since $g(1)=0$, for $x>1$, we have $g(x)<0$, i.e., $0<2 \ln x<x-\frac{1}{x}$.
Therefore, $2 \ln \sqrt{1+\frac{2}{k}}<\sqrt{1+\frac{2}{k}}-\frac{1}{\sqrt{1+\frac{2}{k}}}=\frac{2}{\sqrt{k(k+2)}}$,
i.e., $\ln \left(1+\frac{2}{k}\right)<\frac{2}{\sqrt{k(k+2)}}$
Thus, $\ln ^{2}\left(1+\frac{2}{k}\right)<\frac{4}{k(k+2)}$
Hence, $\sum_{k=1}^{n} \ln ^{2}\left(1+\frac{2}{k}\right)<\sum_{k=1}^{n} \frac{4}{k(k+2)}=2 \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+2}\right)$
$=2\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)<3$.
.20 points

(1) Given $f(x)=p x-\frac{p}{x}-2 \ln x$, we have $f^{\prime}(x)=p+\frac{p}{x^{2}}-\frac{2}{x}$.
To make $f(x)$ a monotonically increasing function in its domain $(0,+\infty)$, we need $f^{\prime}(x) \geq 0$,
i.e., $p x^{2}-2 x+p \geq 0$ holds for all $x \in (0,+\infty)$.
Thus, $p \geq \frac{2 x}{x^{2}+1}$. Note that: $\frac{2 x}{x^{2}+1} \leq \frac{2 x}{2 x}=1$, with equality when $x=1$,
i.e., $\frac{2 x}{x^{2}+1}$ reaches its maximum value of 1 at $x=1$.
Therefore, $p \geq 1$.
5 points
(2) Solution 1: Note that $g(x)=\frac{2 e}{x}$ is a decreasing function on $[1, e]$,
so $g(x)_{\min }=g(e)=2, g(x)_{\max }=g(1)=2 e$, i.e., $g(x) \in[2,2 e]$.
When $0 < p \leq 1$, $f(x)$ is a decreasing function on $[1, e]$, and $f(x)_{\min }=f(e)=p\left(e-\frac{1}{e}\right)-2 \ln e \leq 2$, which does not satisfy the condition.
When $p > 1$, $f(x)$ is an increasing function on $[1, e]$, and we need $f(x)_{\max }=f(e)=p\left(e-\frac{1}{e}\right)-2 \ln e > 2$, solving this gives $p>\frac{4 e}{e^{2}-1}$.
In summary, the range of $p$ is $\left(\frac{4 e}{e^{2}-1},+\infty\right)$.

Solution 2: The original proposition is equivalent to $f(x)-g(x)>0$ having a solution on $[1, e]$,
Let $F(x)=f(x)-g(x)=p x-\frac{p}{x}-2 \ln x-\frac{2 e}{x}$.
Since $F^{\prime}(x)=p+\frac{p}{x^{2}}-\frac{2}{x}+\frac{2 e}{x^{2}}=\frac{p x^{2}+p+2(e-x)}{x^{2}}>0$,
$F(x)$ is an increasing function,
Thus, $F(x)_{\max }=F(e)>0$, solving this gives $p>\frac{4 e}{e^{2}-1}$.
Therefore, the range of $p$ is $\left(\frac{4 e}{e^{2}-1},+\infty\right)$.