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Let $J_{\nu}(z)$ denote the Bessel function of the first kind. Then, according to this page, we have the recurrence relation

$$\displaystyle J_{\nu}(z) = \frac{2(\nu + 1)}{z}J_{\nu + 1}(z) - J_{\nu + 2}.$$

Taking $\nu = 1$, we can inductively express $J_{1}$ in the following ways:

$$\displaystyle J_{1}(z) = \frac{4}{z}J_{2}(z) - J_{3}(z)$$ $$\displaystyle = \frac{4}{z}\left( \frac{6}{z}J_{3}(z) - J_{4}(z) \right) - \left( \frac{8}{z}J_{4}(z) - J_{5}(z) \right)$$ $$\displaystyle = \ ...$$

Thus, my question is: can we write

$$\displaystyle J_{1}(z) = \sum_{n = 1}^{\infty}\frac{a_{n}J_{\nu_n}(z)}{z^{n}}$$

for some real coefficients $a_i$ and indices $\nu_i$, or something similar?

This question seems to suggest that we can, for large $z$.

Equation 10.17.3 here also seems relevant, except without the Bessel functions appearing in the summands. I am particularly interested if it is possible to come up with a series starting with $z^{-1}$ rather than $z^0$.

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    With decreasing powers of $z$ ... you may want it to converge near $+\infty$. (Jack showed it cannot converge near $0$.) I am guessing even that is not possible, because $J_1(z)$ oscillates. But I have no proof. – 2017-02-16
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    @GEdgar: we are dealing with entire functions: how can we have convergence in a left neighbourhood of $+\infty$ without having convergence in a right neighbourhood of $0$, too? – 2017-02-16
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    So, maybe it converges in $|z|>5$ or something. For example $\sum 1/z^n$ converges in $|z| > 1$. – 2017-02-16

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That is clearly not possible: $J_n(x)$ is an entire function, hence $x=0$ is a regular point. Assuming $$ J_1(x) = \sum_{n\geq 1}\frac{a_n\,J_{\nu_n}(x)}{x^n} \tag{1}$$ (with $\nu_n\in\mathbb{N}$) holds, we have that $x=0$ is a singularity for $J_1(x)$, contradiction.
However, we have $$ \mathcal{L}\left(J_n(x)\right) = \frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^n}\tag{2} $$ that can be used to produce similar identities through geometric series expansion and $\mathcal{L}^{-1}$.