Thread: limits

find the limits of

(1+2+3+...+n) /n^2

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.