\\[tex]\\a_{n+1}=\frac{3}{2(n+1)+3}=\frac{3}{2n+5} \\a_{n+1}-a_n=\frac{3}{2n+5}-\frac{3}{2n+3}=\frac{3(2n+3)-3(2n+5)}{(2n+5)(2n+3)}= \\\frac{6n+9-6n-15}{(2n+5)(2n+3)}=\frac{-6}{(2n+5)(2n+3)}<0\implies \{a_n\} \ malejacy[/tex]
an = 3 / (2n + 3)
a(n+1) = 3 / [2(n + 1) + 3] = 3 / [2n + 2 + 3] = 3 / (2n + 5)
a(n+1) - an = 3 / (2n + 5) - 3 / (2n + 3) = [ 3(2n + 3) - 3(2n + 5) ] / [(2n + 3)(2n + 5) ] =
[ 6n + 9 - 6n - 15) ] / [(2n + 3)(2n + 5) ] = (-6) / [(2n + 3)(2n + 5) ] < 0
ciąg jest ciągiem malejącym