116 Let the increasing sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=1,4 a_{n+1}=5 a_{n}+\sqrt{9 a_{n}^{2}+16}(n \geqslant$ $1, n \in \mathbf{N}^{*}$ ).
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}}<2$.
16 (1) Method 1: From $a_{1}=1, 4 a_{n+1}=5 a_{n}+\sqrt{9 a_{n}^{2}+16}(n \geqslant 1, n \in \mathbf{N}^{*})$, 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
$$
a_{n}=\frac{1+4+4^{2}+\cdots+4^{n-1}}{2^{n-1}}=\frac{4^{n}-1}{3 \times 2^{n-1}}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right).
$$

Proof: When $n=1$, $a_{1}=\frac{2}{3}\left(2-\frac{1}{2}\right)=1$ is clearly true;
Assume when $n=k$, $a_{k}=\frac{2}{3}\left(2^{k}-\frac{1}{2^{k}}\right)$ holds, when $n=k+1$,
$$
\begin{aligned}
5 a_{k}+\sqrt{9 a_{k}^{2}+16} & =\frac{10}{3}\left(2^{k}-\frac{1}{2^{k}}\right)+\sqrt{4\left(2^{k}-\frac{1}{2^{k}}\right)^{2}+16} \\
& =\frac{10}{3}\left(2^{k}-\frac{1}{2^{k}}\right)+2\left(2^{k}+\frac{1}{2^{k}}\right) \\
& =\frac{16}{3} \times 2^{k}-\frac{4}{3} \times \frac{1}{2^{k}} \\
& =\frac{8}{3}\left(2^{k+1}-\frac{1}{2^{k+1}}\right) \\
& =4 a_{k+1}
\end{aligned}
$$

holds, so the general term formula is $a_{n}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right)$.
Method 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 $x^{2}-\frac{5}{2} x+1=0$, with roots $x_{1}=2, x_{2}=\frac{1}{2}$, so $a_{n}=a_{1} x_{1}^{n}+a_{2} x_{2}^{n}$, substituting $a_{1}=1, a_{2}=\frac{5}{2}$, we get
$$
\left\{\begin{array}{l}
2 a_{1}+\frac{1}{2} a_{2}=1, \\
4 a_{1}+\frac{1}{4} a_{2}=\frac{5}{2},
\end{array}\right.
$$

solving, we get
$$
\left\{\begin{array}{l}
a_{1}=\frac{2}{3}, \\
a_{2}=-\frac{2}{3},
\end{array}\right.
$$

the general term formula is $a_{n}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right)$.
(2) Let $S_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}}$, from
$$
2^{n}-\frac{1}{2^{n}}=2\left(2^{n-1}-\frac{1}{2^{n-1}}\right)>2\left(2^{n-1}-\frac{1}{2^{n-1}}\right)(n \geqslant 2)
$$

we get $a_{n}>2 a_{n-1}, \frac{1}{a_{n}}<\frac{1}{2} \cdot \frac{1}{a_{n-1}}(n \geqslant 2)$,
$$
\begin{aligned}
S_{n} & =\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}} \\
& <\frac{1}{a_{1}}+\frac{1}{2}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n-1}}\right) \\
& =\frac{1}{a_{1}}+\frac{1}{2}\left(S_{n}-\frac{1}{a_{n}}\right) \\
& <\frac{1}{a_{1}}+\frac{1}{2} \cdot S_{n}.
\end{aligned}
$$

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