When [tex]n=1[/tex], we have
[tex]\dfrac n{2n+1}=\dfrac1{2\cdot1+1}=\dfrac13[/tex]
When [tex]n=2[/tex], we have
[tex]\dfrac n{2n+1}=\dfrac2{2\cdot2+1}=\dfrac25[/tex]
and so on. The series is then
[tex]\displaystyle\sum_{n=1}^6\frac n{2n+1}=\frac13+\frac25+\frac37+\frac49+\frac5{11}+\frac6{13}[/tex]
