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Dec 31, 2010 at 18:57 comment added Suvrit Maybe this is going the wrong way round, but another possible bound can be obtained by using $I_\nu(z) = (z/2)e^z M(0.5+\nu; 1+2\nu; 2z) / \Gamma(\nu+1)$, where $M$ is Kummer's function. Now, you can invoke Stirling's approximation for $\Gamma$, and one of the many approximations for $M$.
Dec 31, 2010 at 18:45 vote accept user11936
Dec 31, 2010 at 17:54 answer added John D. Cook timeline score: 1
Dec 31, 2010 at 17:31 comment added Zsbán Ambrus The Abramowitz-Stegun Handbook has some asymptotics that may help you.
Dec 31, 2010 at 17:04 history asked user11936 CC BY-SA 2.5