$a_n=\dfrac23\left(2^n-\dfrac{1}{2^n}\right)$

由题意得$$a_2=\dfrac52,a_3=\dfrac{21}{4},a_4=\dfrac{85}{8}.$$因为 $5=1+4,21=1+4+4^2,85=1+4+4^4+4^3$，猜想$$\begin{split}a_n&=\dfrac{1+4+4^2+\cdots+4^{n-1}}{2^{n-1}}\\ &=\dfrac{4^n-1}{3\cdot2^{n-1}}=\dfrac23\left(2^n-\dfrac{1}{2^n}\right),\end{split}$$下面用数学归纳法证明：
归纳基础当 $n=1$ 时，$$a_1=\dfrac23\left(2-\dfrac12\right)=1$$显然成立；
递推证明当 $n=k$ 时，$$a_k=\dfrac23\left(2^k-\dfrac{1}{2^k}\right)$$成立．
当 $n=k+1$ 时，根据归纳假设，有\[\begin{split}5a_k+\sqrt{9a_k^2+16}&=\dfrac{10}{3}\left(2^k-\dfrac{1}{2^k}\right)+\sqrt{4\left(2^k-\dfrac{1}{2^k}\right)^2+16}\\&=\dfrac{10}{3}\left(2^k-\dfrac{1}{2^k}\right)+2\left(2^k+\dfrac{1}{2^k}\right)\\&=\dfrac83\left(2^{k+1}-\dfrac{1}{2^{k+1}}\right),\end{split}\]所以$$a_{k+1}=\dfrac 23\left(2^{k+1}-\dfrac {1}{2^{k+1}}\right),$$即对 $n=k+1$ 也成立．
因此通项公式为 $a_n=\dfrac23\left(2^n-\dfrac{1}{2^n}\right)$．

略

设 $S_n=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n}$．
因为$$2^n-\dfrac{1}{2^n}=2\left(2^{n-1}-\dfrac{1}{2^{n+1}}\right)>2\left(2^{n-1}-\dfrac{1}{2^{n-1}}\right)(n\geqslant2),$$所以$$a_n>2a_{n-1} , \dfrac{1}{a_n}<\dfrac12\cdot\dfrac{1}{a_{n-1}}(n\geqslant2),$$故\[\begin{split}S_n&=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n}\\&<\dfrac{1}{a_1}+\dfrac12\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_{n-1}}\right)\\&=\dfrac{1}{a_1}+\dfrac12\left(S_n-\dfrac{1}{a_n}\right)\\&<\dfrac{1}{a_1}+\dfrac12\cdot S_n,\end{split}\]因此 $S_n<\dfrac{2}{a_1}=2$．