When the Helmholtz equation is separated in spherical coordinates the radial equation has the form
$$r^2\frac{d^2R}{dr^2}+2r\frac{dR}{dr}+[k^2r^2-n(n+1)]R=0.\ \ \ \ \ \mbox{(1)}$$
The equation (1) looks similar to Bessel’s equation.  If we use the transformation $R(kr)=\frac{Z(kr)}{(kr)^{1/2}}$, (1) turns into Bessel’s equation
$$r^2\frac{d^2Z}{dr^2}+r\frac{dZ}{dr}+\left[k^2r^2-\left(n+\frac{1}{2}\right)^2\right]Z=0.\ \ \ \ \ \mbox{(2)}$$
Hence $Z(kr)=J_{n+\frac{1}{2}}(x)$, Bessel function of order $n+\frac{1}{2}$ where $n$ is an integer.

Spherical Bessel Functions: Spherical Bessel functions of the first kind and the second kind are defined by
\begin{align*}
j_n(x)&:=\sqrt{\frac{\pi}{2x}}J_{n+\frac{1}{2}}(x),\\
n_n(x)&:=\sqrt{\frac{\pi}{2x}}N_{n+\frac{1}{2}}(x)=(-1)^{n+1}\sqrt{\frac{\pi}{2x}}J_{-n-\frac{1}{2}}(x).
\end{align*}
Spherical Bessel functions $j_n(kr)$ and $n_n(kr)$ are two linearly independent solutions of the equation (1).

One can obtain power series representations of $j_n(x)$ and $n_n(x)$ using Legendre Duplication Formula

$$z!\left(z+\frac{1}{2}\right)!=2^{-2z-1}\pi^{1/2}(2z+1)!$$ from $$J_{n+\frac{1}{2}}(x)=\sum_{s=0}^\infty\frac{(-1)^s}{s!\left(s+n+\frac{1}{2}\right)!}\left(\frac{x}{2}\right)^{2s+n+\frac{1}{2}}:$$
\begin{align*}
j_n(x)&=2^nx^n\sum_{s=0}^\infty\frac{(-1)^s(s+n)!}{s!(2s+2n+1)!}x^{2s},\\
n_n(x)&=(-1)^{n+1}\frac{2^n\pi^{1/2}}{x^{n+1}}\sum_{s=0}^\infty\frac{(-1)^s}{s!\left(s-n-\frac{1}{2}\right)!}\left(\frac{x}{2}\right)^{2s}\\
&=\frac{(-1)^{n+1}}{2^nx^{n+1}}\sum_{s=0}^\infty\frac{(-1)^s(s-n)!}{s!(2s-2n)!}x^{2s}.
\end{align*}
From these power series representations, we obtain
\begin{align*}
j_0(x)&=\frac{\sin x}{x}\left(=\sum_{s=0}^\infty\frac{(-1)^s}{(2s+1)!}x^{2s}\right)\\
n_0(x)&=-\frac{\cos x}{x}\\
j_1(x)&=\frac{\sin x}{x^2}-\frac{\cos x}{x}\\
n_1(x)&=-\frac{\cos x}{x^2}-\frac{\sin x}{x}.
\end{align*}
Orthogonality: Recall the orthogonality of Bessel functions
$$\int_0^aJ_\nu\left(\frac{\alpha_{\nu p}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu q}}{a}\rho\right)\rho d\rho=\frac{a^2}{2}[J_{\nu+1}(\alpha_{\nu p})]^2\delta_{pq}$$ as discussed here. By a substitution, we obtain the orthogonality of spherical Bessel functions
$$\int_0^aj_n\left(\frac{\alpha_{np}}{a}\rho\right)j_n\left(\frac{\alpha_{nq}}{a}\rho\right)\rho^2 d\rho=\frac{a^3}{2}[j_{n+1}(\alpha_{np})]^2\delta_{pq},$$ where $\alpha_{np}$ and $\alpha_{nq}$ are roots of $j_n$.

Example: [Particle in a Sphere]

Let us consider a particle inside a sphere with radius $a$. The wave function that describes the state of the particle satisfies Schrödinger equation
$$-\frac{\hbar^2}{2m}\nabla^2\psi=E\psi\ \ \ \ \ \mbox{(3)}$$
with boundary conditions:
\begin{align*}
&\psi(r\leq a)\ \mbox{is finite},\\
&\psi(a)=0.
\end{align*}
This corresponds to a potential $V=0$, $r\leq a$ and $V=\infty$, $r>a$. Here $m$ is the mass of the particle, $\hbar=\frac{h}{2\pi}$ is the reduced Planck constant (also called Dirac constant).
Note that (3) is the Helmholtz equation $\nabla^2\psi+k^2\psi=0$ with $k^2=\frac{2mE}{\hbar^2}$, whose radial part satisfies
$$\frac{d^2R}{dr^2}+\frac{2}{r}\frac{dR}{dr}+\left[k^2-\frac{n(n+1)}{r^2}\right]R=0.$$ Now we determine the minimum energy (zero-point energy) $E_{\mbox{min}}$. Since any angular dependence would increase the energy, we take $n=0$. The solution $R$ is given by
$$R(kr)=Aj_0(kr)+Bn_0(kr).$$ Since $n_0(kr)\rightarrow\infty$ at the origin, $B=0$. From the boundary condition $\psi(a)=0$, $R(a)=0$, i.e. $j_0(ka)=0$. Thus $ka=\frac{2mE}{\hbar}a=\alpha$ is a root of $j_0(x)$.The smallest $\alpha$ is the first zero of $j_0(x)$, $\alpha=\pi$. Therefore,
\begin{align*}
E_{\mbox{min}}&=\frac{\hbar^2\alpha^2}{2ma^2}\\
&=\frac{\hbar^2\pi^2}{2ma^2}\\
&=\frac{h^2}{8ma^2},
\end{align*}
where $h$ is the Planck constant. This means that for any finite sphere, the particle will have a positive minimum energy (or zero-point energy).