Rewrite $\sum _ { i = 1 } ^ { n } \frac { 7 i ^ { 3 } } { n ^ { 8 } }$ as a rational function S(n)and find $\lim _ { n \rightarrow \infty } S ( n )$ .
A) $S ( n ) = \frac { 7 ( n + 1 ) ^ { 2 } } { 4 n ^ { 6 } } , \lim _ { n \rightarrow \infty } S ( n ) = 0$
B) $S ( n ) = \frac { 7 n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } , \lim _ { n \rightarrow \infty } S ( n ) = 7$
C) $S ( n ) = \frac { 7 ( n + 1 ) ^ { 2 } } { 4 n ^ { 6 } } , \text { the limit does not exist }$
D) $S ( n ) = \frac { 7 ( n + 1 ) ^ { 2 } } { 4 n ^ { 2 } } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 7 } { 4 }$
E)$S ( n ) = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 28 } , \lim _ { n \rightarrow \infty } S ( n ) = 0$

Question

Quizplus · Accepted Answer

The sum can be rewritten using the formula for the sum of cubes, and the limit as $n$ approaches infinity of the resulting rational function is 0 due to the higher power of $n$ in the denominator compared to the numerator.

Rewrite ∑i=1n7i3n8\sum _ { i = 1 } ^ { n } \frac { 7 i ^ { 3 } } { n ^ { 8 } }∑i=1n​n87i3​

Rewrite $\sum _ { i = 1 } ^ { n } \frac { 7 i ^ { 3 } } { n ^ { 8 } }$