$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$ fine the limits :
$$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$$
My Try :
$$\lim_{n\to \infty} \frac{3}{n}(1+\frac{1}{\sqrt{1+\frac{3}{n}}}+\frac{1}{\sqrt{1+\frac{6}{n}}}+...+\frac{1}{\sqrt{1+\frac{3(n-1)}{n}}})\\$$
Now ?

Note that $$\lim _{ n\to \infty  } \frac { 3 }{ n } \left( 1+\frac { 1 }{ \sqrt { 1+\frac { 3 }{ n }  }  } +\frac { 1 }{ \sqrt { 1+\frac { 6 }{ n }  }  } +...+\frac { 1 }{ \sqrt { 1+\frac { 3(n-1) }{ n }  }  }  \right) =\\ =3\lim _{ n\to \infty  } \sum _{ k=1 }^{ n }{ \frac { 1 }{ n } \left( \frac { 1 }{ \sqrt { 1+\frac { 3\left( k-1 \right)  }{ n }  }  }  \right)  } =3\int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { 1+3x }  }  } =\int _{ 0 }^{ 1 }{ \frac { d\left( 1+3x \right)  }{ \sqrt { 1+3x }  }  } =\\ =2$$
