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?
 A: Note that
$$
0\le 1-\cos\frac{1}{n}=2\sin^2\frac{1}{2n}\le 2\cdot\left(\frac{1}{2n}\right)^2=\frac{1}{2n^2}.
$$
We have used above that $$1-\cos (2x)=2\sin^2 x,$$ and also that $0 \le \sin x\le x$, whenever $x\in [0,\pi/2]$. 
A: A small addendum. It has already been pointed out that the series is clearly convergent, since $1-\cos\frac{1}{n}$ behaves like $\frac{C}{n^2}$ for large values of $n$, hence asymptotic comparison and the p-test are conclusive. In explicit terms,
$$ \sum_{n\geq 1}\left(1-\cos\tfrac{1}{n}\right)=2\sum_{n\geq 1}\sin^2\tfrac{1}{2n} $$
can be efficiently approximated through Bhaskara's $\cos y\approx \frac{\pi^2-4y^2}{\pi^2+y^2}$ and Poisson's summation formula, leading to
$$ S=\sum_{n\geq 1}\left(1-\cos\tfrac{1}{n}\right)\approx \frac{5}{e^2-1} $$
whose relative approximation error is already less than one part in two hundreds. An exact representation is provided by
$$ S = \sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m)!}\zeta(2m)=\sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m)!(2m-1)!}\int_{0}^{+\infty}\frac{z^{2m-1}}{e^z-1}\,dz\\= -\int_{0}^{+\infty}\frac{\text{ber}_1(2\sqrt{z})+\text{bei}_1(2\sqrt{z})}{(e^z-1)\sqrt{2z}}\,dz$$
where $\text{ber}$ and $\text{bei}$ are Kelvin functions. The last integrand function is concentrated in a right neighbourhood of the origin, where it behaves like $\frac{1}{2}\exp\left(-\frac{x}{2}-\frac{x^2}{18}\right)$. The error of the resulting approximation
$$ S\approx \frac{3}{2} e^{9/8} \sqrt{\frac{\pi }{2}} \text{Erfc}\left[\frac{3}{2 \sqrt{2}}\right]$$
essentially has the same magnitude of the previous one.
A: We can apply the Limit Test with  $\rho =2$,
$$\lim_{k \to \infty} k^2\left( 1 - \cos{\frac{1}{k}}\right) = \lim_{u \to 0}\frac{1 - \cos{u}}{u^2} = \frac{1}{2}$$
and thus the series converges absolutely.
A: I am a student too so I hope my answer is correct.
$1-\cos(1/n)=\cos(0)-\cos(1/n)=-2\sin(1/2n)\sin(-1/2n)=2\sin^2(1/2n)$
Now as $\sin(1/2n)<1/2n \; \forall n \in \mathbb{N} \Rightarrow \sin^2(1/2n)<1/4n^2$
Hence $ |\sum_{n=1}^{\infty }(1-\cos(1/n))| \leq \sum_{n=1}^{\infty }|(1-\cos(1/n))|=\sum_{n=1}^{\infty }|2\sin^2(1/2n)| \leq \sum_{n=1}^{\infty }|(2/4n^2)|< \infty $. Because we all ready know that $\sum_{n=1}^{\infty }1/n^2$ converge
A: 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
