Answer
$$f(x)=\cosh x=\frac{e^{x}+e^{-x}}{2}\\
=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...$$
Convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows
$$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$
Now by the comparison with the function $f(x)=\cosh x=\frac{e^{x}+e^{-x}}{2}$, we have the Maclaurin series as follows
$$f(x)=\cosh x=\frac{e^{x}+e^{-x}}{2}\\
=\frac{1}{2}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\
+\frac{1}{2}(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\
=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence, the Maclaurin series for $f(x)=\cosh x$ is also convergent for any value of $x$.
