Convergence or divergence of $\sum_{k=1}^{\infty} \left(1-\cos\frac{1}{k}\right)$ [duplicate]

Hints:

• Note that $1-\cos\frac{1}{k} = 2 \cdot \sin^{2}\frac{1}{2k}$

• $\sin^{2}\frac{1}{2k} < \frac{1}{4k^2}$.

In reply to Michael's comment:

• Note that $\cos(A+B) = \cos(A)\cdot \cos(B) - \sin(A)\cdot \sin(B)$.

• So $\cos(2x) = \cos^{2}(x)-\sin^{2}(x) = 1-\sin^{2}(x)-\sin^{2}(x)=1-2\sin^{2}(x)$.

• So from above $2\:\sin^{2}(x)= 1 -\cos(2x)$. Put $x = \frac{1}{2k}$.

$1-\cos\left(\frac 1k\right)=\int^{1/k}_0\sin tdt$ hence $0\leq 1-\cos\left(\frac 1k\right)\leq \int_0^{1/k}tdt=\frac 1{2k^2}$, and we can conclude, since $\sum_{k\geq 1}\frac 1{k^2}$ is convergent.

We know that $\cos x = 1-\frac{x^2}{2} + o(x^3)$ (this is $\cos x$ Taylor expansion near zero), so:

$$1-\cos\left(\frac{1}{k}\right) = 1-\left(1-\frac{\left(\frac{1}{k}\right)^2}{2} + o\left(\frac{1}{k^2}\right) \right)=\frac{\left(\frac{1}{k}\right)^2}{2} + o\left(\frac{1}{k^2}\right)$$

Notice that:

$$\lim_{k\to\infty}\frac{\frac{\left(\frac{1}{k}\right)^2}{2} + o\left(\frac{1}{k^2}\right)}{\frac{1}{k^2}}=\frac{1}{2}$$

Since $\sum \frac{1}{k^2}$ converges, and since $\left(1-\cos\left(\frac{1}{k}\right)\right)>0$ for all natural $k$, we'll conclude from the limit comparison test that $\sum \left(1-\cos\frac{1}{k} \right)$ converges.

Look at the function $$\frac{1+\cos\left(\frac{1}{x}\right)}{\frac{1}{x^2}}$$when $\,\,x\to\infty\,\,$ and apply L'Hospital twice...or even once only if you already know $$\lim_{x\to 0}\frac{\sin x}{x}=1$$