Convergence of an infinite series involving conjugates I have the infinite series 
$$\sum_{n=1}^\infty \left(1-\cos\frac{1}{n}\right) $$
I have to find if it converges or not, and I know I have to use the conjugate find it. So I get
$$\frac{\sin^2\left( \frac{1}{n}\right)}{1+\cos\left( \frac{1}{n}\right)}$$
But now I have no idea what to do (or how that helped)
I'm in high school and this is a section of the unit on series, a section that focuses on integral comparison test, ratio test, and the $n$th root test. Please keep your explanations simple and thanks for the help!
 A: You could also add.
$$1-\cos\left(\frac{1}{n}\right)=\cos(0)-\cos\left(\frac{1}{n}\right)=2\sin^2\left(\frac{1}{2n}\right)$$
And then compare $$0\leq2\sin^2\left(\frac{1}{2n}\right)\leq\frac{1}{n^2}$$
This tell us that $$\sum_{n=1}^{\infty}\left(1-\cos\left(\frac{1}{n}\right)\right)\leq\sum_{n=1}^{\infty}\frac{1}{n^2}\leq1+\sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1+\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)$$
The latter series telescopes to $1$.
A: You just need to exploit the inequality $|\sin x|\leq|x|$:
$$0\leq \sum_{n=1}^{+\infty}\left(1-\cos\frac{1}{n}\right) = 2\sum_{n=1}^{+\infty}\sin^2\frac{1}{2n}\leq 2\sum_{n=1}^{+\infty}\frac{1}{4n^2}=\frac{\pi^2}{12}.$$
Moreover, since:
$$1-\cos x = \sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m)!}x^{2m} $$
we have:
$$ \sum_{n=1}^{+\infty}\left(1-\cos\frac{1}{n}\right)=\sum_{m\geq 1}\frac{(-1)^{m+1}\zeta(2m)}{(2m)!} $$
hence the real value of the series is between $\frac{\pi^2}{12}$ and $\frac{\pi^2}{12}-\frac{\pi^4}{2160}$, so quite close to $\color{red}{\frac{7}{9}}$.
A: If you have covered the Limit Comparison Test, you could use the facts that 
$\hspace{.3 in}\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges and
$\hspace{.27 in}\displaystyle\lim_{n\to\infty}\frac{1-\cos\frac{1}{n}}{\frac{1}{n^2}}=\lim_{t\to 0^{+}}\frac{1-\cos t}{t^2}=\lim_{t\to 0^{+}}\frac{\sin t}{2t}=\frac{1}{2}\lim_{t\to 0^{+}}\frac{\sin t}{t}=\frac{1}{2}\cdot 1=\frac{1}{2}$,
$\;\;\;\;$so $\displaystyle\sum_{n=1}^{\infty}\left(1-\cos\frac{1}{n}\right)$ converges.
