granica ciągu iza: a) a_n = 3+6+9+...+3n / (2n+3)(7n−1) b) a_n = 1+2+3+...+n / 4+8+12+...+4n

Janek191: a_n = ( 3 + 6 + 9 + ... + 3n) /[ ( 2n + 3)*(7n − 1)] b₁ = 3 b_n = 3n r = 6 − 3 = 3 S_n = 0,5 *{ b₁ + b_n]*n = 0,5*[ 3 + 3n]*n = 1,5 *(1 + n)*n więc a_n = [ 1,5 n *(1 + n)]/ [ (2n + 3)*( 7n − 1)] = [ 1,5 *( 1/n + 1)]/[ (2 + 3/n)*( 7 − 1/n)] więc lim a_n = 1,5/14 = 3/28 n→+∞ bo 1/n → 0, 3/n → 0, 1/n → 0, gdy n → +∞

Janek191: b) 1 + 2 + 3 + ... + n = 0,5 n*(n + 1) 4 + 8 + 12 + ... + 4n = 4*(1 + 2 + 3 + ... + n ) więc a_n = 1/4 czyli lim a_n = 1/4 n→+∞