Define the undefined..

$\displaystyle{\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +.....+\frac { 1 }{ n+5n } } \\ }$

$\displaystyle{=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ 5n }{ \frac { 1 }{ n+r } } } }$

$\displaystyle{=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \sum _{ r=1 }^{ 5n }{ \frac { 1 }{ 1+\frac { r }{ n } } } } }$

$\displaystyle{=\int _{ 0 }^{ 5 }{ \frac { 1 }{ 1+x } dx } =\ln { 6 } -\ln { 1 } =ln6\quad }$

$\displaystyle{So,\quad a+b=6+1=7}$

Let us use the notation $H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}.$ The number $H_n$ is called the $n^{th}$ - harmonic number. Then $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+5n}=H_{6n}-H_{n}=H_{6n}-\ln{6n}-H_{n}+\ln{n}+\ln{6n}-\ln{n}=(H_{6n}-\ln{6n})-(H_{n}-\ln{n})+\ln{6}$

Now using that $\lim_{n\rightarrow \infty}(H_n-\ln n)=\gamma,$ where $\gamma$ is the Euler–Mascheroni constant, we obtain that $\lim_{n\rightarrow \infty} (H_{6n}-H_{n})=\ln{6}.$ So that, the answer, $1+6=\boxed{7}.$