4 The sum $S _ { N }$ is defined by $S _ { N } = \sum _ { n = 1 } ^ { N } n ^ { 5 }$. Using the identity

$$\left( n + \frac { 1 } { 2 } \right) ^ { 6 } - \left( n - \frac { 1 } { 2 } \right) ^ { 6 } \equiv 6 n ^ { 5 } + 5 n ^ { 3 } + \frac { 3 } { 8 } n$$

find $S _ { N }$ in terms of $N$. [You need not simplify your result.]

Hence find $\lim _ { N \rightarrow \infty } N ^ { - \lambda } S _ { N }$, for each of the two cases\\
(i) $\lambda = 6$,\\
(ii) $\lambda > 6$.

\hfill \mbox{\textit{CAIE FP1 2010 Q4}}