Challenging +1.2 This question requires pairing terms in an alternating series and using the sum of cubes formula, which is a standard technique for Further Maths students. The limit calculation is straightforward once the expression is found. While it involves multiple steps and some algebraic manipulation, the approach is methodical and follows a well-established pattern for alternating series, making it moderately above average difficulty but not requiring deep insight.
5 Let
$$S _ { N } = \sum _ { n = 1 } ^ { N } ( - 1 ) ^ { n - 1 } n ^ { 3 }$$
Find \(S _ { 2 N }\) in terms of \(N\), simplifying your answer as far as possible.
Hence write down an expression for \(S _ { 2 N + 1 }\) and find the limit, as \(N \rightarrow \infty\), of \(\frac { S _ { 2 N + 1 } } { N ^ { 3 } }\).
5 Let
$$S _ { N } = \sum _ { n = 1 } ^ { N } ( - 1 ) ^ { n - 1 } n ^ { 3 }$$
Find $S _ { 2 N }$ in terms of $N$, simplifying your answer as far as possible.
Hence write down an expression for $S _ { 2 N + 1 }$ and find the limit, as $N \rightarrow \infty$, of $\frac { S _ { 2 N + 1 } } { N ^ { 3 } }$.
\hfill \mbox{\textit{CAIE FP1 2004 Q5 [7]}}