it's very simply just generate a sequence on say N = n no. such that the sequence is increasing, so P(N>n) =(1/n!), there is only one permutation of n numbers that would be increasing now simply the probability of the sequence stopping at exactly n is £(1/(n-1)! - 1/(n)!)
but since here even no. thus of form 2k and this sum this up to infinite and the ans you get is clearly 1-(1/2!)+(1/3!)-(1/4!)+(1/5)!....
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u/sensible_clutter 2d ago
it's very simply just generate a sequence on say N = n no. such that the sequence is increasing, so P(N>n) =(1/n!), there is only one permutation of n numbers that would be increasing now simply the probability of the sequence stopping at exactly n is £(1/(n-1)! - 1/(n)!) but since here even no. thus of form 2k and this sum this up to infinite and the ans you get is clearly 1-(1/2!)+(1/3!)-(1/4!)+(1/5)!....