Good afternoon,
I am currently studying on Bessel functions using Watson's Treatise on the Theory of Bessel functions.
In chapter 3, the series defining the Bessel function of first kind for arbitrary $\nu$, is given by
$$J_\nu(z) = \sum_{m=0}^\infty \frac{(-1)^m (z/2)^{\nu + 2m}}{m! \Gamma(\nu + m + 1)}.$$
In paragraph 3.13, uniform convergence on any bounded domain of values of $z$ and $\nu$ is shown. Let me cite
For, when $|\nu| \le N$ and $|z| \le \Delta$, the test ratio for this series is $$\left|\frac{-\frac{1}{4} z^2}{m(\nu + m)}\right| \le \frac{\frac{1}{4} \Delta^2}{m(m-N)} < 1,$$ whenever $m$ is taken to be greater than the positive root of the equation $$m^2 - mN -\frac{1}{4}\Delta^2 = 0.$$ This choice of $m$ being independent of $\nu$ and $z$, the result stated [about uniform convergence] follows from the test of Weierstrass.
(from: Watson: A Treatise on the Theory of Bessel functions, 2nd ed., Cambridge University Press 1944, p. 44)
The inequalities are not a problem. I assume that the Weierstrass M test is applied here, which I am aware of. However, I do not understand why the "test ratio" is considered here, and how this is sufficient to apply the Weierstrass test. Usually, one has to bound the functions itself, not the ratio.
Maybe someone could help me out here. It would be greatly appreciated. Thank you in advance.
Best regards.
Watson is showing for a series $\sum_{m \geqslant 0}f_m(\nu,z)$ that for all $(\nu,z) \in [-N,N] \times [-\Delta,\Delta]$ and all $m \geqslant M$, where $M$ is independent of $\nu$ and $z$, we have
$$\frac{|f_{m+1}(\nu,z)|}{|f_m(\nu,z)|} \leqslant a_m < 1$$
Since $a_m$ has no dependence on $\nu$ and $z$, and $a_m \to 0$ as $m \to \infty$, there exists $r \in (0,1)$ and $Q \in \mathbb{N}$, independent of $\nu$ and $z$ and with $Q > M$, such that for all $m \geqslant Q$
$$\frac{|f_{m+1}(\nu,z)|}{|f_m(\nu,z)|} \leqslant r$$
By induction, we can show that for all $m > Q$ and for all $(\nu,z) \in [-N,N] \times [-\Delta,\Delta]$, we have
$$|f_m(\nu,z)| < \frac{|f_Q(\nu,z)|}{r^Q}r^m$$
Since $f_Q(\nu,z) /r^Q$ is bounded on $[-N,N] \times [-\Delta,\Delta]$ it follows that
$$|f_m(\nu,z)| < Cr^m,$$
and, by the Weierstrass test, the series is uniformly convergent.