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Does the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$ converge..? Can anyone please give me a simple proof..

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Just use Taylor expansion for $\sin (1/n)$ as $n \to \infty$ – Cortizol Feb 26 '13 at 20:07
up vote 6 down vote accepted

Consider that

$$\frac{1}{n} - \sin{\left(\frac{1}{n}\right)} \sim \frac{1}{6 n^3}$$

as $n \rightarrow \infty$. Then use the comparison test.

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In case you don't know the power series but do know that $\sin'(t) = \cos(t)$ and $\sin(t) \le t$ for $t \ge 0$:

$x-\sin(x) = \int_0^x (1-\cos(t))dt$.

$\cos(2t) = \cos^2(t)-\sin^2(t) = 1-2\sin^2(t)$ so $1-\cos(t) = 2\sin^2(t/2) \le t^2/2$ since $|\sin(t)| \le |t|$.

Therefore $x-\sin(x) \le \int_0^x (t^2/2)dt = x^3/6$ and the remainder of the proof goes through as before.

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