a_n=\frac{2n+1}{4n+3}
a_{n+1}=\frac{2(n+1)+1}{4(n+1)+3}=\frac{2n+2+1}{4n+4+3}=\frac{2n+3}{4n+7}
a_{n+1}-a_n=\frac{2n+3}{4n+7}-\frac{2n+1}{4n+3}=\frac{(2n+3)(4n+3)-(2n+1)(4n+7)}{(4n+7)(4n+3)}=
=\frac{8n^2+6n+12n+9-(8n^2+14n+4n+7)}{16n^2+12n+28n+21}=
\frac{8n^2+18n+9-8n^2-18n-7}{16n^2+40n+21}=\frac{2}{16n^2+40n+21}
a_{n+1}-a_n>0| > Ciąg rosnący