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I saw this approximation posted in a different thread

Convergence of the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$..?.

I am very curious as to how one would come up with this approximation. My intuition tells me it has something to do with power series, but I cannot find a way to show this. I would like to learn more about how to get this result so I can apply it to the other trigonometric functions.

giobrach
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Jungleshrimp
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    With a Taylor expansion – Dalamar Apr 28 '18 at 18:15
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    You are right. Taylor series: $\sin(x)=x-x^3/3!+o(x^4)$. Therefore $\frac{1}{n}-\sin(1/n)=1/(3!n^3)+o(n^{-4})$. It follows that $6n^3(1/n-\sin(1/n))=1+o(n^{-4})\to 1$ as $n\to\infty$. –  Apr 28 '18 at 18:16

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By Taylor series, you know that $$\sin(x) = \sum_{n=0}^m (-1)^{n}\frac{x^{2n+1}}{(2n+1)!} +\mathtt o(x^{2m+2})$$ as $x \to 0$. The notation $\mathtt o(x^{2m+2})$ means that there exists a function $q$ that is infinitesimal as $x \to 0$ such that $$\sin(x) = S_m(x) + x^{2m+2}q(x), $$ where $S_m(x)$ is the $(2m+1)$-th (and $(2m+2)$-th) Taylor polynomial of $\sin$, which we wrote in the first equation. The sequence $1/n$ converges to $0$, so this is the expansion you are interested in; we get, for $m = 1$, $$\sin\left(\frac 1 n \right) = \frac 1 n - \frac{1}{3! n^3} + \mathtt o\left(\frac 1 {n^4}\right)$$ so $$\frac 1 n - \sin\left(\frac 1 n \right) = \frac{1}{3! n^3} + \mathtt o\left(\frac 1 {n^4}\right), $$ as we wanted.

giobrach
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