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

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joemath314159Best ResponseYou've already chosen the best response.2
You can factor the 1+n^3 out of the denominator, then use the fact that:\[\sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
is it 1/3 ? @joemath314159
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
yes thats correct :)
 one year ago
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