Show that \(\tanh ( \ln n ) = \frac { n ^ { 2 } - 1 } { n ^ { 2 } + 1 }\).
It is given that, for non-negative integers \(n , I _ { n } = \int _ { 0 } ^ { \ln 2 } \tanh ^ { n } u \mathrm {~d} u\).
Show that \(I _ { n } - I _ { n - 2 } = - \frac { 1 } { n - 1 } \left( \frac { 3 } { 5 } \right) ^ { n - 1 }\), for \(n \geqslant 2\).
Find the value of \(I _ { 3 }\), giving your answer in the form \(a + \ln b\), where \(a\) and \(b\) are constants.
Use the method of differences on the result of part (ii) to find the sum of the infinite series
$$\frac { 1 } { 2 } \left( \frac { 3 } { 5 } \right) ^ { 2 } + \frac { 1 } { 4 } \left( \frac { 3 } { 5 } \right) ^ { 4 } + \frac { 1 } { 6 } \left( \frac { 3 } { 5 } \right) ^ { 6 } + \ldots .$$