Just Divide and Conquer.

Calculus Level 5

$\large\ \lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \lambda { k }^{ 4 } + 2{ k }^{ 3 } + { k }^{ 2 } + k + 1 }{ { 3n }^{ 5 } + { n }^{ 2 } + n + 5k } } } = \frac { 1 }{ 3 }$

Find $\lambda$ if the equation above holds true.