## Yahoo! $Sn = \sum_{k=1}^{n}\frac{ k^2 }{ 1+n^3 }$ , then $\lim_{n \rightarrow \infty }Sn = ?$ one year ago one year ago

1. joemath314159

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

2. Yahoo!

Yup..i got it...thxxx

3. Yahoo!

is it 1/3 ? @joemath314159

4. joemath314159

yes thats correct :)