My approach is slightly more rigorous than the OP's, as it is not generally justified to manipulate 'infinite tower fractions' as if it is finite.

*However mine is much slower.*

We have the recursion $$T_n=n+\frac{T_{n+2}}{n+1}$$ and the OP asks for $T_1$.

First, define $V_n=T_{2n+1}$. Then, $$V_n=2n+1+\frac{V_{n+1}}{2n+2}\implies V_{n}=-2n(2n-1)+2nV_{n-1}$$ and we want to find $V_0$.

Recall the extremely useful formula for first order recursion:
$$f_n=\alpha_n+\beta_n f_{n-1}\implies f_n=f_0\prod^n_{i=1}\beta_i+\sum^{n-1}_{k=0}\alpha_{n-k}\prod^k_{j=1}\beta_{n-j+1}$$

Applying it gives
$$\begin{align}
V_n
&=V_0\prod^n_{i=1}(2i)-\sum^{n-1}_{k=0}(2(n-k)-1)(2(n-k))\prod^k_{j=1}2(n-j+1) \\
&=V_02^n n!-\sum^{n}_{k=1}(2k-1)(2k)\prod^{n-k}_{j=1}2(n-j+1) \\
&=V_02^n n!-\sum^{n}_{k=1}(2k-1)(2k)\prod^{n}_{j=k+1}2j \\
&=V_02^n n!-\sum^{n}_{k=1}(2k-1)(2k)\cdot 2^{n-k}\frac{n!}{k!} \\
\frac{V_n}{2^nn!}&=V_0-\sum^{n}_{k=1}\frac{4k^2-2k}{2^kk!} \qquad{(\star)} \\
\end{align}
$$

From the recurrence relation, it is apparent that $V_n=O(n)$. Therefore, taking limit $n\to\infty$ on both sides of $(\star)$,
$$V_0-\sum^{\infty}_{k=1}\frac{4k^2-2k}{2^kk!}=0$$

We can apply known formulae to evaluate this sum very quickly, but I would like to demonstrate that closed form of this sum can be derived purely by elementary methods:
$$\begin{align}
\sum^{\infty}_{k=1}\frac{4k^2-2k}{2^kk!}
&=\sum^{\infty}_{k=1}\frac{4k-2}{2^k(k-1)!} \\
&=\sum^{\infty}_{k=0}\frac{4k+2}{2^{k+1} k!} \\
&=2\sum^{\infty}_{k=0}\frac{k}{2^k k!}+\sum^{\infty}_{k=0}\frac{1}{2^k k!} \\
&=2\sum^{\infty}_{k=1}\frac{k}{2^k k!}+\sqrt e \\
&=2\sum^{\infty}_{k=1}\frac{1}{2^k (k-1)!}+\sqrt e \\
&=2\sum^{\infty}_{k=0}\frac{1}{2^{k+1} k!}+\sqrt e \\
&=\sum^{\infty}_{k=0}\frac{1}{2^{k} k!}+\sqrt e \\
V_0&=2\sqrt e \qquad{\blacksquare}\\
\end{align}
$$

*p.s. I have not seen this problem before.*

*p.p.s. We made the same careless mistake :)*