Question

\[Sn = \sum_{k=1}^{n}\frac{ k^2 }{ 1+n^3 }\] , then \[\lim_{n \rightarrow \infty }Sn = ?\]

Answer 1

You can factor the 1+n^3 out of the denominator, then use the fact that:\[\sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]

Answer 2

Yup..i got it...thxxx

Answer 3

is it 1/3 ? @joemath314159

Answer 4

yes thats correct :)