Determine if a series converges, cosine, Taylor How do I determine if the series
$$ \sum_{k=1}^{\infty} 1- \cos \frac{1}{k} $$
converges?
I got a hint that I should use Taylor expansion, but I cant figure out how it would be applied correctly here.
 A: Since$$\lim_{k\to\infty}\frac{1-\cos\left(\frac1k\right)}{\frac1{k^2}}=\frac12$$and since $\sum_{k=1}^\infty\frac1{k^2}$ converges, your series converges too.
A: The mean value theorem and $\sin x \leq x$ for $x\geq 0$ is also enough:
$$0\leq 1-\cos\frac{1}{k} \stackrel{0<\xi_k <\frac{1}{k}}{\leq}\frac{1}{k} \sin \xi_k \leq \frac{1}{k^2}$$
A: Hint:
Use Taylor Expansion to get $1-\cos{\dfrac1k}=\dfrac1{2k^2}+o\left(\dfrac1{k^2}\right)$ then conclude using Riemann criterion.
A: Actually you do not need Taylor series. If $x\in(0,1]$ we have
$$ 0\leq 1-\cos(x) = 2\sin^2\frac{x}{2} \leq 2\left(\frac{x}{2}\right)^2 = \frac{x^2}{2},$$
hence 
$$ 0\leq \sum_{k\geq 1}\left(1-\cos\tfrac{1}{k}\right)\leq \frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2}=\frac{\pi^2}{12}.$$

To compute the actual value of the series is a different task, and a quite challenging one. From
$$ 1-\cos\tfrac{1}{k} = \sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n)! k^{2n}} $$
we get
$$ S=\sum_{k\geq 1}\left(1-\cos\tfrac{1}{k}\right) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n)!}\zeta(2n)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n)!(2n-1)!}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\,dx $$
from which
$$ S = -\sqrt{2}\int_{0}^{+\infty}\frac{\text{bei}_1(2x)+\text{ber}_1(2x)}{e^{x^2}-1}\,dx $$
with $\text{bei}$ and $\text{ber}$ being Kelvin functions. The last integrand function converges to zero extremely fast (it belongs to the Schwartz space $\mathscr{S}(\mathbb{R})$), so Newton-Cotes formulas or Gaussian quadrature schemes are perfectly suited for evaluating the integral over $[0,M]$ with arbitrary accuracy. We have $S\approx 0.778759$.
