Official solution
\[{{{e^x} + {e^{ - x}}} \over 2}\]
\[ = 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........\]
Putting \[x = {1 \over 2}\] we get
\[1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......\]
\[\infty = {{{e^{{1 \over 2}}} + {e^{{{ - 1} \over 2}}}} \over 2} = {{\sqrt e + {1 \over {\sqrt e }}} \over 2}\]
\[ = {{e + 1} \over {2\sqrt e }}\]