Soit \(n\in\Nn^*\).

\[\begin{align*}\ensuremath \sum_{k=1}^{n}\sin\frac{k}{n^2}&=\mbox{Im}(\sum_{k=1}^{n}e^{ik/n^2})=\mbox{Im}\left(e^{i/n^2}\frac{1-e^{ni/n^2}}{1-e^{i/n^2}}\right) =\mbox{Im}\left(e^{i(1+\frac{n}{2}-\frac{1}{2})/n^2}\frac{\sin\frac{1}{2n}}{\sin\frac{1}{2n^2}}\right) =\frac{\sin\frac{n+1}{2n^2}\sin\frac{1}{2n}}{\sin\frac{1}{2n^2}}\\ &\underset{n\rightarrow+\infty}{=}(\frac{1}{2n}+\frac{1}{2n^2}+o(\frac{1}{n^2}))(\frac{1}{2n}+o(\frac{1}{n^2}))(\frac{1}{2n^2}+o(\frac{1}{n^3}))^{-1}\\ &=(1+\frac{1}{n}+o(\frac{1}{n}))(\frac{1}{2}+o(\frac{1}{n}))(1+o(\frac{1}{n}))^{-1}=\frac{1}{2}+\frac{1}{2n} +o(\frac{1}{n}), \end{align*}\]

(on peut aussi partir de l’encadrement \(\frac{k}{n^2}-\frac{k^3}{6n^6}\leq\sin\frac{k}{n^2}\leq\frac{k}{n^2}\)).

exo7 5458