Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1 .$$
Show that \(\frac { \mathrm { d } } { \mathrm { dx } } \left( \tan ^ { - 1 } ( \sinh x ) \right) = \operatorname { sech } x\).
Sketch the graph of \(y = \operatorname { sechx }\), stating the equation of the asymptote.
By considering a suitable set of \(n\) rectangles of unit width, use your sketch to show that
$$\sum _ { r = 1 } ^ { n } \operatorname { sechr } < \tan ^ { - 1 } ( \operatorname { sinhn } )$$
Hence state an upper bound, in terms of \(\pi\), for \(\sum _ { r = 1 } ^ { \infty }\) sech \(r\).