What can we say about the series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$

Question

Series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$

$$\sum_{n=1}^{\infty}n\cdot(1-\cos(\frac{a}{n}))=\sum_{n=1}^{\infty}n\cdot[1- \{1-\frac{a^2}{2!\cdot n^2} + O(\frac{1}{n^4})\}] = \sum_{n=1}^{\infty} \frac{a^2}{2n} + O(\frac{1}{n^3})$$

Can we conclude with Limit comparison test with $\sum \frac1n$, that the given series is divergent?

Also can we say anything about its oscillatory nature?

Answer 1

Yes we have that

$$n\cdot\left(1-\cos\left(\frac{a}{n}\right)\right)=\frac{a^2}{n}\frac{\left(1-\cos\left(\frac{a}{n}\right)\right)}{\frac{a^2}{n^2}}\sim \frac{a^2}{2n}$$

then the series diverges by limit comparison test.

Answer 2

Another way, using $\frac{2x}{\pi} \le \sin(x) \le x$ for $0 \le x \le \pi/2$:

$\begin{array}\\ \sum_{n=1}^{m}n(1-\cos(\frac{a}{n})) &=\sum_{n=1}^{m}n(2\sin^2(\frac{a}{2n}))\\ &=2\sum_{n=1}^{m}n\sin^2(\frac{a}{2n})\\ &\ge 2\sum_{n=1}^{m}n(\frac{2a}{2\pi n})^2 \qquad\text{for }\frac{a}{2n} \le \frac{\pi}{2} \text{ or }n \ge \frac{a}{\pi} \\ &\ge 2\sum_{n=\lceil \frac{a}{\pi} \rceil}^{m}n(\frac{a}{\pi n})^2\\ &=\frac{2a^2}{\pi^2}\sum_{n=\lceil \frac{a}{\pi}\rceil}^{m}\frac1{n}\\ \end{array} $

and this diverges by comparison with $\sum \frac1{n}$.