Rain Man

First off, I'd like to say that I liked this problem quite a lot, which is what has compelled me to write this solution. Well done sir.

$\lim _{ n\rightarrow \infty }{ \sum _{ k=0 }^{ 3n }{ \frac { 1 }{ n+k } } = } \lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \sum _{ k=0 }^{ 3n }{ \frac { 1 }{ 1+\frac { k }{ n } } } }$

Then I let $\frac{k}{n} = x_k$

$\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \sum _{ k=0 }^{ 3n }{ \frac { 1 }{ 1+{ x }_{ k } } } =\int _{ 0 }^{ 3 }{ \frac { 1 }{ 1+x } =\ln { 4 } } }$

For more information on this method of finding the limit, search "Playing with Integrals" in the Brilliant search.

let $H_n= nth\quad harmonic$ then the expression equals $\lim_{n\rightarrow\infty} (H_{4n})-\lim_{n\rightarrow\infty} (H_{n-1})$ $=\lim_{n\rightarrow\infty} (H_{4n}-\ln(4n))-\lim_{n\rightarrow\infty} (H_{n-1}-\ln(n-1))+\lim_{n\rightarrow\infty} (\ln(4n)-\ln(n-1))$ we use the fact that $\lim_{n\rightarrow\infty} (H_{n}-\ln(n))=\gamma$ where $\gamma$ is the Euler–Mascheroni constant. (http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant) the limit translates to $\gamma-\gamma+\ln(4)+\ln(\lim_{n\rightarrow\infty}(\dfrac{n}{n-1}))=\boxed{\ln(4)}$