15. Given $f(x)=(\sqrt{2}+\sqrt{x})^{2}(x \geqslant 0)$. In the sequence $\left\{a_{n}\right\}\left(a_{n}>0\right)$, $a_{1}=2$, and the sum of the first $n$ terms of this sequence $S_{n}\left(n \in \mathbf{N}_{+}\right)$ satisfies $S_{n}=f\left(S_{n-1}\right)$ for all positive integers $n$ greater than 1.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If $b_{n}=\frac{a_{n+1}^{2}+a_{n}^{2}}{2 a_{n+1} a_{n}}\left(n \in \mathbf{N}_{+}\right)$, prove that $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+b_{3}+\cdots+b_{n}-\right.$ $n)=1$
15. (1) Since $S_{n}=\left(\sqrt{2}+\sqrt{S_{n-1}}\right)^{2}$, and $S_{n}>0$, we have $\sqrt{S_{n}}=\sqrt{S_{n-1}}+\sqrt{2}$, and $S_{1}=\sqrt{2}$. Therefore, $\sqrt{S_{n}}=\sqrt{2} \cdot n$, which means $S_{n}=2 n^{2}(n \geqslant 2)$. And $a_{n}=S_{n}-S_{n-1}=4 n-2, a_{1}=2$, so
$$
a_{n}=4 n-2(n \geqslant 1) \text {; }
$$
(2) From (1), we know $b_{n}=\frac{\left(a_{n+1}-a_{n}\right)^{2}}{2 a_{n+1} a_{n}}+1=\frac{2}{(2 n+1)(2 n-1)}+1=\frac{1}{2 n-1}-\frac{1}{2 n+1}+1$, so
$$
b_{1}+b_{2}+\cdots+b_{n}=1-\frac{1}{2 n+1}+n, b_{1}+b_{2}+\cdots+b_{n}-n=1-\frac{1}{2 n+1} .
$$

Therefore, $\lim \left(b_{1}+b_{2}+\cdots+b_{n}-n\right)=1$.