Mathematics
It is amazing how 2 divergent series and sequence converge to give Euler's constant:
As n goes to infinity we have (1+1/2+1/3+1/4+...+1/n)-log(n)=0.577.
It is amazing how 2 divergent series and sequence converge to give Euler's constant:
As n goes to infinity we have (1+1/2+1/3+1/4+...+1/n)-log(n)=0.577.
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