Bessel function at large order I'm trying to expand a modified Bessel function such that
$$K_n \left(\sqrt{n} \left(a_0 + a_1 \frac{1}{n} + a_2 \frac{1}{n^2} + \ldots \right)\right) = A(n) \left( b_0 + b_1 \frac{1}{n} + b_2 \frac{1}{n^2} + \ldots \right) $$
in the large $n$ limit and $A(n)$ is some function of $n$ that contains diverging part. Is such expansion possible? How would one obtain the coefficients $b_0, b_1, \ldots$ in terms of $a_0, a_1, \ldots$?
I would appreciate any hint or reference. Thank you.
 A: Unless I misunderstand, there's something missing here.  It looks to me like $K_n(\sqrt{n})$ blows up rather rapidly as $n \to \infty$.  For example, 
for $n=10$ I get approximately $1413.798936$, for $n = 20$ approximately $4.796177691 \times 10^9$.
A: Partial answer. Assume $a_0>0$. Substituting
$$
z = \frac{{a_0 }}{{\nu ^{1/2} }} + \frac{{a_1 }}{{\nu ^{3/2} }} + \frac{{a_2 }}{{\nu ^{5/2} }} +  \ldots 
$$
into the standard uniform asymptotic expansion of $K_\nu(\nu z)$ and expanding the functions of $z$ appearing in the expansion about $z=0$, yields, after some computation,
$$
K_\nu\!\left( {\nu ^{1/2}\! \left( {a_0  + \frac{{a_1 }}{\nu } + \frac{{a_2 }}{{\nu ^2 }} +  \ldots } \right)} \right) = \sqrt {\frac{\pi }{{2\nu }}} \left( {\frac{{2\sqrt \nu  }}{{a_0 {\rm e}}}} \right)^\nu  \exp\! \bigg( { - \frac{{a_0^2 }}{4} - \frac{{a_1 }}{{a_0 }}} \bigg)\!\left( 1 + \mathcal{O}\!\left( {\frac{1}{\nu }} \right) \right)
$$
as $\nu\to+\infty$. The coefficients $a_n$ with $n\ge 2$ will affect the error term only.
The same technique can be used to answer the question. The coefficients $b_n$ may be computed one-by-one with laborious work, but I cannot see any simple way to get a general formula for them.
