Establish the formulae
tanz=n=1∑∞(−1)n−122n(22n−1)b2n(2n)!z2n−1
sinz1=z1+n=1∑∞(−1)n−12(22n−1−1)b2n(2n)!z2n−1
here the series on the right-hand sides converge absolutely, the first for ∣z∣<2π and the second for ∣z∣<π (express tanz and 1/sin2z as linear combinations of cotz and cot2z). Deduce that the numbers 2n22n(22n−1)b2n are integers. (Use the following lemma: in two absolutely convergent series ∑n=0∞αnn!zn, ∑n=0∞βnn!zn the coefficients αn and βn are integers, then, in their product written in the form ∑n=0∞γnn!zn, the γn are integers.)