= Solution
The <hyperbolic cosine> has the absolutely convergent expansion $\cosh(1/2)=\sum_{k\ge0}1/[4^k(2k)!]$. Suppose its value were a rational $a/b$, with $b>0$. For sufficiently large $N$, $b\mid(2N)!$. Multiplying by $4^N(2N)!$ makes the assumed value an integer, and every term of the partial sum through $k=N$ is also an integer. Their difference is the positive remainder
$$
R_N=\sum_{j=1}^\infty\frac{(2N)!}{4^j(2N+2j)!}.
$$
Let $q=[4(2N+1)(2N+2)]^{-1}$. Every new pair of factorial factors is at least $(2N+1)(2N+2)$, so $0<R_N\le\sum_{j\ge1}q^j=q/(1-q)<1$ for $N\ge1$. No integer lies strictly between zero and one. This contradiction proves \b[$\cosh(1/2)$ is irrational]. The argument uses a finite integer partial sum and a controlled tail; it does not infer irrationality merely from an infinite expansion.
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