$\text { Rewrite } \sum_{i=1}^{n} \frac{9 i^{3}}{n^{5}} \text { as a rational function } S(n) \text { and find } \lim _{n \rightarrow \infty} S(n) \text {. }$

A)$S(n)=\frac{9(n+1)^{2}}{4 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{9}{4}$
B)$S(n)=\frac{9 n^{2}(n+1)^{2}}{4}, \lim _{n \rightarrow \infty} S(n)=9$
C)$S(n)=\frac{n^{2}(n+1)^{2}}{36}, \lim _{n \rightarrow \infty} S(n)=0$
D)$S(n)=\frac{9(n+1)^{2}}{4 n^{3}}, \lim _{n \rightarrow \infty} S(n)=0$
E) $S(n)=\frac{9(n+1)^{2}}{4 n^{3}}, \text { the limit does not exist }$

Question

$\text { Rewrite } \sum_{i=1}^{n} \frac{9 i^{3}}{n^{5}} \text { as a rational function } S(n) \text { and find } \lim _{n \rightarrow \infty} S(n) \text {. }$

A)$S(n)=\frac{9(n+1)^{2}}{4 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{9}{4}$
B)$S(n)=\frac{9 n^{2}(n+1)^{2}}{4}, \lim _{n \rightarrow \infty} S(n)=9$
C)$S(n)=\frac{n^{2}(n+1)^{2}}{36}, \lim _{n \rightarrow \infty} S(n)=0$
D)$S(n)=\frac{9(n+1)^{2}}{4 n^{3}}, \lim _{n \rightarrow \infty} S(n)=0$
E) $S(n)=\frac{9(n+1)^{2}}{4 n^{3}}, \text { the limit does not exist }$

Quizplus · Accepted Answer

The Answer of $\text { Rewrite } \sum_{i=1}^{n} \frac{9 i^{3}}{n^{5}} \text { as a rational function } S(n) \text { and find } \lim _{n \rightarrow \infty} S(n) \text {. }$

A)$S(n)=\frac{9(n+1)^{2}}{4 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{9}{4}$
B)$S(n)=\frac{9 n^{2}(n+1)^{2}}{4}, \lim _{n \rightarrow \infty} S(n)=9$
C)$S(n)=\frac{n^{2}(n+1)^{2}}{36}, \lim _{n \rightarrow \infty} S(n)=0$
D)$S(n)=\frac{9(n+1)^{2}}{4 n^{3}}, \lim _{n \rightarrow \infty} S(n)=0$
E) $S(n)=\frac{9(n+1)^{2}}{4 n^{3}}, \text { the limit does not exist }$