Local Uniform Convergence of a series of modified Bessel functions of the first kind

Question

We know that $$ \sum_{k=1}^{\infty} I_k(x)=\frac{1}{2}(e^{x}-I_0(x)) $$ But how can we show that this convergence is locally uniform? In particular how can we show $$ \sum_{k=n}^{\infty} e^{-x}I_k(x) $$ converges locally uniformly to $0$? I have tried finding some nice bound for $e^{-x}I_k(x)$ such that we end up with something that looks geometric, but I have been unable to get any such bound. Any other methods or suitable bounds would be appreciated.

Answer 1

By $(10.32.2)$, we have $$ {\rm e}^{ - x} I_k (x) = \frac{1}{{\sqrt {2\pi } }}\left( {\frac{x}{2}} \right)^k \frac{1}{{\Gamma \!\left( {k + \frac{1}{2}} \right)}}\int_0^\pi {{\rm e}^{ - x(1 - \cos t)} (\sin t)^{2k} {\rm d}t} $$ for any $x\ge 0$ and $k\ge 0$. Accordingly, $$ \left| {{\rm e}^{ - x} I_k (x)} \right| \le \sqrt {\frac{\pi }{2}} \left( {\frac{x}{2}} \right)^k \frac{1}{{\Gamma \!\left( {k + \frac{1}{2}} \right)}} \le \sqrt {\frac{\pi }{2}} \frac{x}{2}\frac{1}{{(k - 1)!}}\left( {\frac{x}{2}} \right)^{k - 1} $$ for any $x\ge 0$ and $k\ge 2$. This gives, for $n\ge 2$, $$ \left| {\sum\limits_{k = n}^\infty {{\rm e}^{ - x} I_k (x)} } \right| \le \sqrt {\frac{\pi }{2}} \frac{x}{2}\sum\limits_{k = n - 1}^\infty {\frac{1}{{k!}}\left( {\frac{x}{2}} \right)^k } , $$ and it is well known that the power series for $\exp(x/2)$ converges uniformly on compact intervals of the positive reals.