16. Let the increasing sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1,4 a_{n+1}=5 a_{n}+\sqrt{9 a_{n}^{2}+16}(n \geqslant 1) \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<2$.
16. (1) Solution 1 From the problem, we get
$$
a_{2}=\frac{5}{2}, a_{3}=\frac{21}{4}, a_{4}=\frac{85}{8}.
$$

From $5=1+4, 21=1+4+4^{2}$,
$$
85=1+4+4^{2}+4^{3},
$$

we conjecture: $a_{n}=\frac{1+4+4^{2}+\cdots+4^{n-1}}{2^{n-1}}$
$$
=\frac{4^{n}-1}{3 \times 2^{n-1}}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right).
$$

In fact, when $n=1$, $a_{1}=\frac{2}{3}\left(2-\frac{1}{2}\right)=1$, which is clearly true.
Assume that when $n=k$, $a_{k}=\frac{2}{3}\left(2^{k}-\frac{1}{2^{k}}\right)$ holds.
When $n=k+1$,
$$
\begin{array}{l}
5 a_{k}+\sqrt{9 a_{k}^{2}+16} \\
=\frac{10}{3}\left(2^{k}-\frac{1}{2^{k}}\right)+\sqrt{4\left(2^{k}-\frac{1}{2^{k}}\right)^{2}+16} \\
=\frac{10}{3}\left(2^{k}-\frac{1}{2^{k}}\right)+2\left(2^{k}+\frac{1}{2^{k}}\right) \\
=\frac{8}{3}\left(2^{k+1}-\frac{1}{2^{k+1}}\right)=4 a^{k+1}
\end{array}
$$

holds.
Therefore, the general term formula is $a_{n}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right)$.
Solution 2 From $4 a_{n+1}=5 a_{n}+\sqrt{9 a_{n}^{2}+16}$, we get $a_{n+1}^{2}+a_{n}^{2}-\frac{5}{2} a_{n+1} a_{n}=1$.
Thus, $a_{n}^{2}+a_{n-1}^{2}-\frac{5}{2} a_{n} a_{n-1}=1(n>1)$.
Subtracting the two equations, we get
$$
\left(a_{n+1}-a_{n-1}\right)\left(a_{n+1}+a_{n-1}-\frac{5}{2} a_{n}\right)=0.
$$

Since $\left\{a_{n}\right\}$ is an increasing sequence, we have
$$
a_{n+1}-\frac{5}{2} a_{n}+a_{n-1}=0(n>1).
$$

The characteristic equation is $x^{2}-\frac{5}{2} x+1=0$.
Thus, the characteristic roots are $x_{1}=2, x_{2}=\frac{1}{2}$.
Therefore, $a_{n}=\alpha_{1} x_{1}^{n}+\alpha_{2} x_{2}^{n}$.
Substituting $a_{1}=1, a_{2}=\frac{5}{2}$, we get
$$
\left\{\begin{array}{l}
2 \alpha_{1} + \frac{1}{2} \alpha_{2} = 1, \\
4 \alpha_{1} + \frac{1}{4} \alpha_{2} = \frac{5}{2}
\end{array} \Rightarrow \left\{\begin{array}{c}
\alpha_{1}=\frac{2}{3}, \\
\alpha_{2}=-\frac{2}{3}.
\end{array}\right.\right.
$$

Thus, the general term formula is $a_{n}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right)$.
(2) Let $S_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$.
From $2^{n}-\frac{1}{2^{n}}=2\left(2^{n-1}-\frac{1}{2^{n+1}}\right)$
$>2\left(2^{n-1}-\frac{1}{2^{n-1}}\right)$
$\Rightarrow a_{i}>2 a_{n-1} \Rightarrow \frac{1}{a_{n}}<\frac{1}{2} \cdot \frac{1}{a_{n-1}}$.
Thus, $S_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$
$$
\begin{array}{l}
<\frac{1}{a_{1}}+\frac{1}{2}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n-1}}\right) \\
=\frac{1}{a_{1}}+\frac{1}{2}\left(S_{n}-\frac{1}{a_{n}}\right) \\
<\frac{1}{a_{1}}+\frac{1}{2} S_{n}.
\end{array}
$$

Therefore, $S_{n}<\frac{2}{a_{1}}=2$.