think in a 2nd wayworm speed is < the series of : .01 * ( 1 + 1/n + 1/n*(n-1) + 1/ n*(n-1)*(n-2) ... )
sum : .01 * ( n + 1+1/2 + 1/3 ... + 1/2 + 1/6 + 1/12 ... + 1/6+1/24+1/60 ... ... )
< .01 * ( n + ln(n) + (1-1/n) + (1-1/n)/3 + (1-1/n)/4 ... )
< .01 * ( n + ln(n) + (1-1/n) * ln(n) )
< n
unless worm speed > 1. else will never catch up.
I don't know how do you derive that and what is n stand for
Probably you have misunderstanding in my question.
Apparently, as long as worm speed > 0, the speed of worm relative to ground is increasing at every time instance and approaching the string stretching speed. Since the increasing of worm speed will never stop until reaching string stretching speed, it will definitely catch up after some time.
Apparently, as long as worm speed > 0, the speed of worm relative to ground is increasing at every time instance and approaching the string stretching speed. Since the increasing of worm speed will never stop until reaching string stretching speed, it will definitely catch up after some time.
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