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I've found two ways of proving that
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&leq 2 end{align}
Are there any other ways out there, for proving this?

METHOD 1

Let $kin Bbb{N}$, then

begin{align} f:[ 0&,1]longrightarrow Bbb{R}\&x mapsto
cosleft(dfrac{x}{k}right) end{align}
is continuous. Then, by Mean Value Theorem, there exists $cin [ 0,x]$ such that
begin{align} cosleft(dfrac{x}{k}right)-cosleft(0right) =-dfrac{1}{k}sinleft(dfrac{c}{k}right),(x-0), end{align}
which implies begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right| &=left|sum^{infty}_{k=1}dfrac{x}{k}sinleft(dfrac{c}{k}right)right| leq sum^{infty}_{k=1}dfrac{left|xright|}{k}left|sinleft(dfrac{c}{k}right)right|leq sum^{infty}_{k=1}dfrac{left|xright|}{k}dfrac{left|cright|}{k}\&leq sum^{infty}_{k=1}left(dfrac{left|xright|}{k}right)^2leq sum^{infty}_{k=1}dfrac{1}{k^2}=1+ sum^{infty}_{k=2}dfrac{1}{k^2}\&leq 1+ sum^{infty}_{k=2}dfrac{1}{k(k-1)}\&= 1+ limlimits_{ntoinfty}sum^{n}_{k=2}left(dfrac{1}{k-1}-dfrac{1}{k}right)\&=1+ limlimits_{ntoinfty}left(1-dfrac{1}{n}right)\&=2, end{align}

METHOD 2

Let $xin [0,1]$ be fixed, then
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}dfrac{1}{k}left[-kcosleft(dfrac{x}{k}right)right]^{1}_{0}=sum^{infty}_{k=1}dfrac{1}{k}int^{1}_{0}sinleft(dfrac{x}{k}right)dx \&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx end{align}
The series $sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
begin{align} left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k^2}. end{align}
Hence,
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx=int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx, end{align}
and
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&=left|int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dxright|leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right)right|dx \&leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k^2}dx \&leq 2 end{align}

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

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

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!

Using the Taylor series of the $cos$ function, your sum is equal to the much faster converging $sum_{nge 1} (-1)^{n+1}frac{zeta(2n)}{(2n)!}$. Summing the first two terms, $pi^2/12-pi^4/2160$, gives a value of $approx 0.7773$ with an error of $epsilon_1approx1.388times 10^{-3}$. By contrast, summing two terms of your original sum will give you $approx 0.5821$, this time with a larger error of $epsilon_2 approx .1942$.

To answer your question, we can use a theorem about alternating series, which states that for $S_n=sum_{k=1}^n (-1)^{k+1} a_k$ and $S=lim_{ntoinfty}S_n$, we have that
$$|S-S_n|< a_{n+1}$$
Using the same two terms as above, we can see that $S<pi^6/680400 - pi^4/2160 + pi^2/12approx 0.7788$. Thus,

$$sum_{k=1}^infty 1-cos(1/k) < 0.78 <2$$