Limits of First Kind Bessel Function

Question

If we use the series defintion of the Bessel function (first kind)$$J_0(z)=\sum_{m=0}^{\infty}{\frac{(-1)^m}{(m!)^2}\left(\frac{z}{2}\right)^{2m}}$$ Can you prove that $$\lim_{z\to\infty}{J_0(z)=\lim_{z\to\infty} \sum_{m=0}^{\infty}{\frac{(-1)^m}{(m!)^2}{\left(\frac{z}{2}\right)}^{2m}} = 0}$$ Or can this only be done with the integral representation and the asymptotic expansion here?

Answer 1

This is not an answer but a comment on why this isn't a good idea even if it might be doable. The series representation is a Taylor series around $z = 0$; it's well-suited to computations for small $z$ but there's no reason to expect it to be particularly well-suited to computations for large $z$, given all the signs. This situation already occurs for the simpler series

$$\exp(-z) = \sum_{n=0}^{\infty} (-1)^n \frac{z^n}{n!}$$

where we have $\lim_{z \to \infty} \exp(-z) = 0$; can you tell that the limit is zero from this series? The problem is that for $z$ large a tremendous amount of cancellation needs to be happening among the terms; the largest term occurs when $n \approx z$ and grows like $\exp(z)$, yet the final result ends up converging to zero (and quite rapidly) as $z \to \infty$. This is really not clear at all from the series, and I don't know a direct argument from the series at all. It's just not well-suited to understanding what happens when $z$ is large even if it does technically converge and we should use something else.

Similarly, in this case for $z$ large a tremendous amount of cancellation needs to be happening. The largest term occurs when $m \approx \frac{z}{2}$ and again grows like $\exp(z)$, and again the final result needs to end up converging to zero.