Does the series $ \sum_{n=1}^{\infty}\left( 1-\cos\big(\frac{1}{n} \big) \right)$ converge? I'm having trouble determining whether the series:
$$
\sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right]
$$
converges.
I have tried the root test:
$$\lim_{n\rightarrow\infty}\sqrt[n]{1-\cos\frac{1}{n}}=\lim_{n\rightarrow\infty}\left(1-\cos\frac{1}{n}\right)^{1/n}=\lim_{n\rightarrow\infty}\mathrm{e}^{\frac{\log(1-\cos\frac{1}{n})}{n}}=\mathrm{e}^{\lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n})}{n}}$$
Now by applying the Stolz–Cesàro theorem, that upper limit is equal to:
\begin{align}
\lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})}{(n+1)-n}&=\lim_{n\rightarrow\infty}\left(\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})\right)
\\&=\lim_{n\rightarrow\infty}\log{\frac{1-\cos{\frac{1}{n+1}}}{1-\cos{\frac{1}{n}}}}
\end{align}
Now I'm totally stuck, unless that quotient is actually 1, in which case the limit would be 0, the Root test result would be $\mathrm{e}^0=1$ and all this would have been to no avail.
I'm not sure this method was the best idea, the series sure seems way simpler than that, so probably another method is more appropriate?

We know:
$$\cos t \sim 1-\dfrac{t^2}{2}$$
Hence, for $n \to +\infty$
$$\sum_{n=1}^{+\infty}\left(1-\cos\left(\dfrac{1}{n}\right)\right)\sim \sum_{n=1}^{+\infty}\dfrac{1}{2n^2}=\frac{\pi^2}{12}$$
Thus it converges
