Interesting Limit (16)

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{n \to \infty} n^3\left((n+1){\color{#3D99F6}\sin \frac 1{n+1}} - n {\color{#3D99F6}\sin \frac 1n} \right) & \small \color{#3D99F6} \text{Using Maclaurin series} \\ & = \lim_{n \to \infty} n^3\left((n+1){\color{#3D99F6}\left(\frac 1{n+1} - \frac 1{3!(n+1)^3} + \frac 1{5!(n+1)^5}+\cdots \right)} - n {\color{#3D99F6}\left(\frac 1n - \frac 1{3!n^3} + \frac 1{5!n^5}+\cdots \right)} \right) \\ & = \lim_{n \to \infty} \left( \left(n^3 - \frac {n^3}{3!(n+1)^2} + \frac {n^3}{5!(n+1)^4}+\cdots \right) - \left(n^3 - \frac {n^3}{3!n^2} + \frac {n^3}{5!n^4}+\cdots \right)\right) \\ & = \lim_{n \to \infty} \left(\frac {n^3}{3!n^2} - \frac {n^3}{3!(n+1)^2} - {\color{#D61F06} \cancel{\frac {n^3}{5!n^4}}^0 + \cancel{\frac {n^3}{5!(n+1)^4}}^0 + \cancel{ \frac {n^3}{7!n^6}}^0 - \cancel{\frac {n^3}{7!(n+1)^6}}^0 + \cdots } \right) \\ & = \lim_{n \to \infty} \frac {n^3(2n+1)}{6n^2(n+1)^2} = \lim_{n \to \infty} \frac {2n^2+n}{6n^2 + 12n+6} = \lim_{n \to \infty} \frac {2+\frac 1n}{6 + \frac {12}n + \frac 6{n^2}} = \frac 13 \approx \boxed{0.333} \end{aligned}$

Note that as $k \to \infty$ , we have $\sin\frac{1}{k} = \frac{1}{k} - \frac{1}{6k^3} + o(k^{-4})$ so in particular, $\begin{aligned} (n+1)\sin\frac{1}{n+1} - n\sin\frac{1}{n} &= (n+1)\left(\frac{1}{n+1} - \frac{1}{6(n+1)^3} + o(n^{-4})\right) - n\left(\frac{1}{n} - \frac{1}{6n^3} + o(n^{-4})\right) \\ &= \frac{1}{6n^2} - \frac{1}{6(n+1)^2} + o(n^{-3}) \\ &= \frac{2n + 1}{6n^2(n+1)^2} + o(n^{-3}) \\ &= \frac{1}{3n(n+1)^2} + o(n^{-3}) \end{aligned}$

Therefore $\begin{aligned} \lim_{n\to\infty} n^3\left((n+1)\sin\frac{1}{n+1} - n\sin\frac{1}{n}\right) &= \lim_{n\to\infty} \left(\frac{n^2}{3(n+1)^2} + o(1)\right) \\ &= \frac{1}{3} + 0 \\ &\approx \boxed{0.333333} \end{aligned}$