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find the limits of (1+2+3+...+n) /n^2

Quote: Originally Posted by alexandrabel90 find the limits of (1+2+3+...+n) /n^2 1+2+3+...+n = n(n+1)/2 (1+2+3+...+n) /n^2 = n(n+1)/(2n^2) = (n^2+n)/(2n^2) = 1/2 + 1/(2n) as n goes to infinity, the value of (1+2+3+...+n) /n^2 goes to 1/2.