Solution
= Solution
Multiply the <Maclaurin series> of <hyperbolic cosine> by the <binomial series> for the reciprocal square root:
$$
\cosh x=1+\frac{x^2}{2}+\frac{x^4}{24}+O(x^6),\qquad
(1+x^2)^{-1/2}=1-\frac{x^2}{2}+\frac{3x^4}{8}+O(x^6).
$$
The quadratic terms cancel, while the quartic coefficient is $1/24-1/4+3/8=1/6$. Hence
$$
\boxed{\frac{\cosh x}{\sqrt{1+x^2}}=1+\frac{x^4}{6}+O(x^6).}
$$
The absence of odd powers follows from the function's even symmetry.