Limit of a Sum

Relevant wiki: Riemann Sums

Rewriting the limit:

$\lim_{x\to\infty} \frac{1}{x} \left(\sum_{n=1}^{\infty} \dfrac{1}{3+\frac{n}{x}}\right)$

Using the limit definition of the integral, this is a definite integral such that $b-a=1$ and $f\left(\frac{1}{x}\right) = \dfrac{1}{3+\frac{1}{x}}$ . Therefore, $f\left(x\right) = \dfrac{1}{3+x}$ , $a=0$ and $b=1$ .

$\int_{0}^{1} \dfrac{1}{3+x} dx$

$\ln{4}-\ln{3}$

$\boxed{0.287}$

Quite simple.

$\displaystyle \lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } \left( \frac { 1 }{ 3+\frac { 1 }{ x } } +\frac { 1 }{ 3+\frac { 2 }{ x } } +\cdots +\frac { 1 }{ 3+\frac { x }{ x } } \right) } =\lim _{ x\rightarrow \infty }{ \left( \frac { 1 }{ 3x+1 } +\frac { 1 }{ 3x+2 } +\cdots +\frac { 1 }{ 4x } \right) } \\ =\lim _{ x\rightarrow \infty } \left( { H }_{ 4x }-{ H }_{ 3x } \right) \\ =\lim _{ x\rightarrow \infty } \left( \left( \ln { \left( 4x \right) +\gamma } \right) -\left( \ln { \left( 3x \right) +\gamma } \right) \right) \\ =\lim _{ x\rightarrow \infty } \left( \ln { \left( \frac { 4x }{ 3x } \right) } \right) \\ =\ln { \left( \frac { 4 }{ 3 } \right) }$