Asymptotic behavior of solutions to laguerre's differential equation The Frobenius solution to the homogenous DE (Laguerre's):
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$$xg'' + (2n+2 -x)g' + (\lambda - n -1)g = 0$$
is given by 
$$g(x) = \sum_{k = 0}^\infty a_k \,x^{n+k}$$
where the coefficients are generated by the recurrence:
$$a_0 = 1, \qquad \frac{a_{k+1}}{a_k} = \frac{n+k+1 - \lambda}{2(n+k+1)(k+1)}$$
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The series truncates for certain integer values of $\lambda$. Multiple texts mention that for non-integer values of $\lambda$ the solution grows on the order of $e^x$ at infinity. Unfortunately, this statement is always prefaced by "it can be shown that ..". I can't seem to find a source that demonstrates this. I was hoping someone could point me in the right direction.
 A: In this context "it can be shown that .." is a shortcut meaning that the full proof would requires tiresome calculus and huge background.
$$xg'' + (2n+2 -x)g' + (\lambda - n -1)g = 0$$
The general solution of this form of associated Laguerre differential equation is,
if $\lambda$ is an integer :
$$g(x)= c_1 U(-\lambda + n +1 \:,\; 2n+2\:,\: x)+c_2L_{\lambda - n -1}^{2n+1}(x) $$
or, if $\lambda$ isn't an integer :
$$g(x)=c_1\: U(-\lambda + n +1 \:,\; 2n+2\:,\: x) +c_2\:\: _1F_1(-\lambda + n +1 \:,\; 2n+2\:,\: x) $$ 
$U(a,b,x)$ is the confluent hypergeometric function of second kind. The equivalent at infinity is $x^{-a}$
$$U(-\lambda + n +1 \:,\; 2n+2\:,\: x\to\infty)\:\sim\:x^{\lambda-n-1}$$
This is not the leading term in $g(x)$ for large $x$.
$L_k^h(x)$ is the associated Laguerre polynomial, with $k$ and $h$ integers. This is not the case here, since $\lambda$ is not an integer. 
$_1F_1(a,b,x)$ is confluent hypergeometric function of first kind. The equivalent  is $\frac{\Gamma(b)}{\Gamma(a)}e^x x^{a-b}$ at infinity. 
$$ _1F_1( -\lambda + n +1 \:,\; 2n+2\:,\: x\to\infty)\:\sim\: \frac{\Gamma(2n+2)}{\Gamma(-\lambda + n +1)} x^{-\lambda - n -1}e^x$$
This is the leading term in $g(x)$ at infinity due to the exponential.
$$g(x\to\infty)\:\sim\: c\: x^{-\lambda - n -1}\:e^x$$
Of course, the above argument isn't an autonomous proof because it refers to a background of known properties of confluent hypergeometric functions. 
