Rewrite $\sum _ { i = 1 } ^ { n } \frac { 5 i ^ { 3 } } { n ^ { 5 } }$ as a rational function S(n)and find $\lim_ { n \rightarrow \infty } S ( n )$ .
A) $S ( n ) = \frac { 5 ( n + 1 ) ^ { 2 } } { 4 n ^ { 2 } } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 5 } { 4 }$
B) $S ( n ) = \frac { 5 n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } , \lim _ { n \rightarrow \infty } S ( n ) = 5$
C) $S ( n ) = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 20 } , \lim _ { n \rightarrow \infty } S ( n ) = 0$
D) $S ( n ) = \frac { 5 ( n + 1 ) ^ { 2 } } { 4 n ^ { 3 } } , \lim _ { n \rightarrow \infty } S ( n ) = 0$
E) $S ( n ) = \frac { 5 ( n + 1 ) ^ { 2 } } { 4 n ^ { 3 } }$ the limit does not exist

Question

Rewrite $\sum _ { i = 1 } ^ { n } \frac { 5 i ^ { 3 } } { n ^ { 5 } }$ as a rational function S(n)and find $\lim_ { n \rightarrow \infty }  S ( n )$ .
A) $S ( n ) = \frac { 5 ( n + 1 ) ^ { 2 } } { 4 n ^ { 2 } } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 5 } { 4 }$ 
B) $S ( n ) = \frac { 5 n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } , \lim _ { n \rightarrow \infty } S ( n ) = 5$
C) $S ( n ) = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 20 } , \lim _ { n \rightarrow \infty } S ( n ) = 0$
D) $S ( n ) = \frac { 5 ( n + 1 ) ^ { 2 } } { 4 n ^ { 3 } } , \lim _ { n \rightarrow \infty } S ( n ) = 0$
E) $S ( n ) = \frac { 5 ( n + 1 ) ^ { 2 } } { 4 n ^ { 3 } }$ the limit does not exist

Quizplus · Accepted Answer

The sum can be rewritten using the formula for the sum of cubes, $\sum_{i=1}^{n}i^3 = \left(\frac{n(n+1)}{2}\right)^2$, and dividing by $n^5$ gives $S(n) = \frac{5}{n^5}\left(\frac{n(n+1)}{2}\right)^2 = \frac{5(n+1)^2}{4n^3}$. The limit as $n \rightarrow \infty$ of this expression is 0, because the numerator grows as $n^2$ while the denominator grows as $n^3$, making the fraction approach 0.

Rewrite ∑i=1n5i3n5\sum _ { i = 1 } ^ { n } \frac { 5 i ^ { 3 } } { n ^ { 5 } }∑i=1n​n55i3​

Rewrite $\sum _ { i = 1 } ^ { n } \frac { 5 i ^ { 3 } } { n ^ { 5 } }$