Answer: (a) \(S_N = \frac16\left(N+\frac12\right)^6 - \frac{5}{24}N^2(N+1)^2 - \frac{1}{32}N(N+1) - \frac{1}{384}\)

(b)(i) \(\lim_{N\to\infty} N^{-6}S_N = \frac16\)

(b)(ii) \(\lim_{N\to\infty} N^{-\lambda}S_N = 0 \quad \text{for } \lambda\gt 6\)

Start from the given identity

\(\left(n+\frac12\right)^6-\left(n-\frac12\right)^6 \equiv 6n^5+5n^3+\frac38n.\)

Sum this from \(n=1\) to \(N\). The left-hand side telescopes:

\(\sum_{n=1}^N \left[\left(n+\frac12\right)^6-\left(n-\frac12\right)^6\right]=\left(N+\frac12\right)^6-\left(\frac12\right)^6.\)

On the right-hand side, use \(S_N=\sum_{n=1}^N n^5\), together with

\(\sum_{n=1}^N n^3=\left(\frac{N(N+1)}{2}\right)^2=\frac14N^2(N+1)^2,\qquad \sum_{n=1}^N n=\frac12N(N+1).\)

So

\(\left(N+\frac12\right)^6-\frac{1}{64}=6S_N+5\cdot\frac14N^2(N+1)^2+\frac38\cdot\frac12N(N+1).\)

Hence

\(\left(N+\frac12\right)^6-\frac{1}{64}=6S_N+\frac54N^2(N+1)^2+\frac{3}{16}N(N+1).\)

Rearranging gives

\(S_N=\frac16\left(N+\frac12\right)^6-\frac{5}{24}N^2(N+1)^2-\frac{1}{32}N(N+1)-\frac{1}{384}.\)

Now consider the limit. The leading term of \(S_N\) is \(\frac16N^6\), so \(S_N\sim \frac16N^6\) as \(N\to\infty\).

Therefore:

For \(\lambda=6\),

\(\lim_{N\to\infty}N^{-6}S_N=\frac16.\)

For \(\lambda\gt 6\),

\(N^{-\lambda}S_N\sim \frac16N^{6-\lambda}\to 0,\)

since \(6-\lambda\lt 0\).