<s>[INST] What can we say about the series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$ 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?
 [/INST] 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
$\sum \frac1{n}$.
</s>