# Does this series: $\sum_0^{\infty} (1-n\sin(1/n))$ converge? [closed]

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

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.

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

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.
Without Taylor: Assume $$0 Then $$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.