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

Question

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

METHOD 1

Let $k\in \Bbb{N}$, then

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

METHOD 2

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

Answer 1

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 $< 0.9$).

Answer 2

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!

Answer 3

Using the Taylor series of the $\cos$ function, your sum is equal to the much faster converging $\sum_{n\ge 1} (-1)^{n+1}\frac{\zeta(2n)}{(2n)!}$. Summing the first two terms, $\pi^2/12-\pi^4/2160$, gives a value of $\approx 0.7773$ with an error of $\epsilon_1\approx1.388\times 10^{-3}$. 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 $S=\lim_{n\to\infty}S_n$, we have that $$|S-S_n|< a_{n+1}$$ Using the same two terms as above, we can see that $S<\pi^6/680400 - \pi^4/2160 + \pi^2/12\approx 0.7788$. Thus,

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

Answer 4

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 $$ 2\sum_{k=1}^{\infty} \sin^2\left( \frac{1}{2k} \right) \le \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k^2} \approx 0.822.$$