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The modified Bessel functions of the first kind fulfill the recurrence relation: $$I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{1}$$ due to the integral representation: $$I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx \tag{2}$$ and the cosine addition formulas. A straightforward consequence of $(1)$ ...

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This particular one is $I_1(2)/I_0(2)$ where $I_0$ and $I_1$ are modified Bessel functions of the first kind, of orders $0$ and $1$.

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There is at least one formula to convert any series to a continued fraction provided none of the terms in the series is zero; see this page from NIST. But you already have a continued fraction and you're trying to go in the opposite direction (to find the formula for an equal series). There are also formulas to convert some continued fractions to series, ...

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Your line of thinking is right: In general, any convergent sequence can be converted into a series whose sum has the same limit: if $a_n$ is a sequence that converges to $a$ as $n$ tends to infinity, put $b_0 = a_0, b_1 = a_1 - a_0, \ldots, b_{i+1} = a_{i+1} - a_i, \ldots$; then $a_n = \sum_{i=0}^n b_n$ and the $b_n$ comprise a convergent series whose sum ...

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