Harmonic Limit

First solution (Simple)

$\displaystyle\lim _{ n\rightarrow \infty }{ \displaystyle\sum _{ k=n+1 }^{ 2n }{ \frac { 1 }{ k } } } \\ =\displaystyle\lim _{ n\rightarrow \infty }{ \displaystyle\sum _{ k=1 }^{ n }{ \frac { 1 }{ n+k } } } \\ =\displaystyle\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \displaystyle\sum _{ k=1 }^{ n }{ \frac { 1 }{ 1+\frac { k }{ n } } } } \\ =\displaystyle\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+x } } \\ =\left. \ln { \left( 1+x \right) } \right| _0^1\\ =\ln { 2 }$

Second solution (Elegant)

$\displaystyle\sum _{ k=n+1 }^{ 2n }{ \frac { 1 }{ k } } \\ =\displaystyle\sum _{ k=1 }^{ 2n }{ \frac { 1 }{ k } } -\displaystyle\sum _{ k=1 }^{ n }{ \frac { 1 }{ k } } \\ =\displaystyle\sum _{ k=1 }^{ 2n }{ \frac { 1 }{ k } } -2\displaystyle\sum _{ k=1 }^{ n }{ \frac { 1 }{ 2k } } \\ =\displaystyle\sum _{ k=1 }^{ 2n }{ \frac { { \left( -1 \right) }^{ k+1 } }{ k } } \\ \therefore \displaystyle\lim _{ n\rightarrow \infty }{ \displaystyle\sum _{ k=n+1 }^{ 2n }{ \frac { 1 }{ k } } } \\ =\displaystyle\lim _{ n\rightarrow \infty }{ \displaystyle\sum _{ k=1 }^{ 2n }{ \frac { { \left( -1 \right) }^{ k+1 } }{ k } } } \\ =\displaystyle\sum _{ k=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ k+1 } }{ k } } \\ =\ln { 2 }$

We can rewrite the summation as :-

$\frac{1}{n}\sum_{k=n+1}^{2n} \frac{n}{k}$

We would now like to use Riemann Sums

The lower limit for integration would be

$\lim_{n\to\infty} \frac{n+1}{n} = 1$

The upper limit of integration would be

$\lim_{n\to\infty} \frac{2n}{n} = 2$

So the given summation can be written as :-

$\int_{1}^{2} \frac{1}{x} dx$

$= \ln(2)$