The vector $\mathrm{r}_i$ is between the beads $i$ and $i+1$. Define a local coordinate system so that $x$ is along $\mathrm{r}_i$. The coordinate $y$ is defined so that $\mathrm{r}_{i-1}$, $\mathrm{r}_{i}$ are both on the same plane, and $z$ is normal to this plane. Thus we can write
\begin{align}
\mathrm{r}_{i-1} &= (\ell\cos\theta, -\ell\sin\theta, 0)_i^T \\
\mathrm{r}_i &= (\ell, 0, 0)_i^T \\
\mathrm{r}_{i+1} &= (\ell\cos\theta, -\ell\sin\theta\cos\varphi_{i+1}, \ell\sin\theta\sin\varphi_{i+1})_i^T
\end{align}
Or after working out the full coordinate transformation matrix from the above information,
$$\begin{pmatrix}x \\ y \\ z \end{pmatrix}_i = A_i\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix}_{i+1}$$
where
$$A_i = \begin{pmatrix}
\cos\theta & -\sin \theta & 0 \\
-\sin\theta\cos\varphi_{i+1} & -\cos\theta\cos\varphi_{i+1} & -\sin\varphi_{i+1} \\
\sin\theta\sin\varphi_{i+1} & \cos\theta\sin\varphi_{i+1} & -\cos\varphi_{i+1}
\end{pmatrix}$$
Now
\begin{align}
\langle R^2 \rangle &= \left \langle \left(\sum_{i=1}^n \mathbf{r}_i\right) \cdot \left(\sum_{j=1}^n \mathbf{r}_j\right) \right\rangle = \sum_{i=1}^n \sum_{j=1}^n \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle \\
&= \sum_{i=1}^n \left( \sum_{j=1}^{i-1} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle + \langle \| \mathbf{r}_i \|^2\rangle + \sum_{j=i+1}^{n} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle \right) \\
&= n\ell^2 + 2\sum_{i=1}^n \sum_{j=i+1}^{n} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle
\end{align}
Using the transition matrix and noting that $j>i$,
\begin{align}
\langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle &= \ell^2 \langle(1, 0, 0) A_i \cdots A_{j-1} (1, 0, 0)^T\rangle = \ell^2 \langle A_i \cdots A_{j-1}\rangle_{11} \\
&=\ell^2 (\langle A\rangle^{j-i})_{11}
\end{align}
Where $A$ is any of the matrices $A_i$, for example we could set $A = A_1$.
Thus
$$\langle R^2 \rangle = n\ell^2 + 2\ell^2 \sum_{i=1}^n \sum_{j=i+1}^{n} (\langle A\rangle^{j-i})_{11} = n\ell^2 + 2\ell^2 \left(\sum_{i=1}^n (n-i) \langle A\rangle^{i}\right)_{11}$$
So towards the characteristic ratio:
\begin{align}
\frac{\langle R^2 \rangle}{n\ell^2} &= 1 + \frac{2}{n} \left(n\langle A\rangle(I-\langle A\rangle^{n})(I-\langle A\rangle)^{-1}\right. \\
& \qquad \left. - \langle A\rangle(I-(n+1)\langle A\rangle^{n} + n\langle A\rangle^{n+1})(I-\langle A\rangle)^{-2}\right)_{11} \\
& = \left((I+\langle A\rangle)(I-\langle A\rangle)^{-1} - \frac{2\langle A \rangle}{n}(I-\langle A\rangle^{n})(I-\langle A\rangle)^{-2}\right)_{11}
\end{align}
Correlation between two beads tends to zero as distance goes to infinity, i.e. $\lim_{n\to\infty}\langle A\rangle^{n} = 0$, so
$$C_\infty = \left((I+\langle A\rangle)(I-\langle A\rangle)^{-1}\right)_{11}$$
If the potential is symmetric, by the virtue of sine being odd $\langle \sin\varphi\rangle = 0$, and we have
$$\langle A\rangle = \begin{pmatrix}
\cos\theta & -\sin \theta & 0 \\
-\sin\theta\langle\cos\varphi\rangle & -\cos\theta\langle\cos\varphi\rangle & 0 \\
0 & 0 & -\langle\cos\varphi\rangle
\end{pmatrix}$$
Remembering that for a 3x3 matrix $T$
$$T^{-1} = \frac{1}{\det(T)}\begin{pmatrix}\det\begin{pmatrix}
T_{22} & T_{23} \\
T_{32} & T_{33}
\end{pmatrix}
& \det\begin{pmatrix}
T_{13} & T_{12} \\
T_{33} & T_{32}
\end{pmatrix}
& \det\begin{pmatrix}
T_{12} & T_{13} \\
T_{22} & T_{23}
\end{pmatrix}
\\
\det\begin{pmatrix}
T_{23} & T_{21} \\
T_{33} & T_{31}
\end{pmatrix}
& \det\begin{pmatrix}
T_{11} & T_{13} \\
T_{31} & T_{33}
\end{pmatrix}
& \det\begin{pmatrix}
T_{13} & T_{11} \\
T_{23} & T_{21}
\end{pmatrix}
\\
\det\begin{pmatrix}
T_{21} & T_{22} \\
T_{31} & T_{32}
\end{pmatrix}
& \det\begin{pmatrix}
T_{12} & T_{11} \\
T_{32} & T_{31}
\end{pmatrix}
& \det\begin{pmatrix}
T_{11} & T_{12} \\
T_{21} & T_{22}
\end{pmatrix}
\end{pmatrix}$$
we have
$$C_\infty = \frac{(1+\cos\theta)(1+\cos\theta\langle\cos\varphi\rangle) + \sin\theta \sin\theta\langle\cos\varphi\rangle}{(1-\cos\theta)(1+\cos\theta\langle\cos\varphi\rangle) - \sin\theta \sin\theta\langle\cos\varphi\rangle}$$
Which gives us the final relation
$$C_\infty = \frac{(1+\cos\theta)(1+\langle\cos\varphi\rangle)}{(1-\cos\theta)(1-\langle\cos\varphi\rangle)}$$
