Una curiosa somma infinita: e^(1-1/2 + 1/3 - 1/4 + 1/5 .... ) = 2
Uguale a
e/e^(1/2) * e^(1/3)/e^(1/4) * e^(1/5)/e^(1/6) *..ecc* e^(1/n)/e^(1/n+1) ..
Diventa:
e^ln(2)= 2
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Una curiosa somma infinita: e^(1-1/2 + 1/3 - 1/4 + 1/5 .... ) = 2