Different ways of proving that $|\sum^{\infty}_{k=1}(1-\cos(1/k))|\leq 2 $

Using the Maclaurin series for $\cos$, your sum is $$S = \sum_{k=1}^\infty \sum_{j=1}^\infty \frac{(-1)^{j+1}}{(2j)!\; k^{2j}}$$ This converges absolutely, and $$ |S| \le \sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(2j)!\; k^{2j}} = \sum_{j=1}^\infty \frac{\zeta(2j)}{(2j)!} \le \sum_{j=1}^\infty \frac{\zeta(2)}{(2j)!} =\frac{(\cosh(1)-1) \pi^2}{6} < 2$$ (in fact $< 0.9$).

Note that $$1 - \cos(x) = \cos(0) - \cos(x) = 2 \sin^2\left( \frac{x}2 \right) $$ by the sum to product identities for all $x$. Therefore,

$$ \sum_{k=1}^{\infty} \left(1- \cos\left( \frac{1}k \right) \right) = 2 \sum_{k=1}^{\infty} \sin^2\left( \frac{1}{2k} \right). $$ Now using the inequality $\sin(x) \le x$ for all positive $x$, we get $$ 2\sum_{k=1}^{\infty} \sin^2\left( \frac{1}{2k} \right) \le \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k^2} \approx 0.822.$$

Another method which gives a better bound than $2$ is the following: the inequality $1-\cos\, x \leq \frac {x^{2}} 2$ holds for all real $x$ and $\sum \frac 1 {k^{2}} =\frac {\pi^{2}} 6$. Use the fact that $\frac {\pi^{2}} {12}<2$. [ $1-\cos\, x - \frac {x^{2}} 2$ vanishes at $0$ and its derivative is $\sin\, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!