The function defined by
$$
f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}, \quad \text { where } \quad
c_{n+2}=\frac{2 n-\lambda}{(n+1)(n+2)} c_{n},
$$
can be written as \(f(x)=c_{0} g(x)+c_{1} h(x)\), where \(g\) is even and \(h\) is
odd in \(x .\) Show that \(f(x)\) goes to infinity at least as fast as \(e^{x^{2}}\)
does, i.e., \(\lim _{x \rightarrow \infty} f(x) e^{-x^{2}} \neq 0 .\) Hint:
Consider \(g(x)\) and \(h(x)\) separately and show that
$$
g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}, \quad \text { where } \quad
b_{n+1}=\frac{4 n-\lambda}{(2 n+1)(2 n+2)} b_{n} .
$$
Then concentrate on the ratio \(g(x) / e^{x^{2}}\), where \(g\) and \(e^{x^{2}}\)
are approximated by polynomials of very high degrees. Take the limit of this
ratio as \(x \rightarrow \infty\), and use recursion relations for \(g\) and
\(e^{x^{2}}\). The odd case follows similarly.