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$\sum_0^{\infty} (1-n\sin(1/n))$

I have no idea how to deal with the sum... I know that $n\sin(1/n) \rightarrow 1$ but it's the sum/dif that I don't get how to work with.

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    $\begingroup$ How do you define the $n=0$ term? The $n\ge1$ sum converges because $1-n\sin\tfrac1n\sim\tfrac16n^{-2}$. $\endgroup$
    – J.G.
    Apr 25, 2021 at 10:12
  • $\begingroup$ Yes: $\int_0^{\infty } \left(1-n \sin \left(\frac{1}{n}\right)\right) \, dn=\frac{\pi }{4}$ $\endgroup$
    – Mariusz Iwaniuk
    Apr 25, 2021 at 10:14

2 Answers 2

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You can use Taylor's theorem, $$ \sin(x) = x - \frac{x^3}{3!} \cos (\xi) $$ for $\xi \in (0, x)$ so that for $n \geqslant 1$, $$ 1-n\sin(1/n) = \frac{1}{n^2\cdot 3!} \cos(\xi_n) $$ Where $\xi_n \in (0, 1/n)$. The sum on the right converges because $\sum 1/n^2$ converges and $|\cos \xi_n| \leqslant 1 $ and therefore $$ \sum_{n=1}^\infty 1 - n\sin(1/n) $$ also converges.

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Without Taylor: Assume $0<x<\pi/2.$ Then $x-\sin x = \int_0^x(1-\cos t)\, dt.$ Now $1-\cos t\le 1-\cos^2 t =\sin^2 t \le t^2.$ Thus $\int_0^x(1-\cos t)\, dt\le x^3/3.$ Therefore

$$1-\frac{\sin x}{x}=\frac{x-\sin x}{x} \le \frac{x^3/3}{x} = x^2/3.$$

It follows that $1-n\sin(1/n) = 1-\dfrac{\sin (1/n)}{1/n} \le (1/n)^2/3.$ This implies convergence of the given series.

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