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

Does $$\sum_{k=1}^{\infty} \left(1-\cos\frac{1}{k}\right)$$ converge or diverge?

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## marked as duplicate by Thursday, studiosus, Rick Decker, Carl Mummert, JimmyK4542Sep 16 '14 at 2:35

Try Limit comparison to $1/k^2$ – user31373 Jun 8 '12 at 15:58
The hint given in question is to consider a p-series but I failed to link them together before asking. – Michael Li Jun 8 '12 at 16:13
Is there any resonable way to calculate its limit? – user 1618033 Jun 8 '12 at 18:14

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.

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$\frac{1}{2}f+o(f)$ is not $\sim f$. And are you sure that's the comparison test? – anon Jun 12 '12 at 7:16
Maybe it's a matter of the $\sim$ notation. Is it ok now? I don't see the problem. – Amihai Zivan Jun 12 '12 at 7:24
@anon: This is a perfectly reasonable use of the limit comparison test. – Erick Wong Jun 12 '12 at 7:26
@ErickWong It appears I was confusing "limit comparison test" with "comparison test." – anon Jun 12 '12 at 7:27
Thanks for your post. This method is rather more obvious than the above ones. – Michael Li Jun 12 '12 at 8:58

Hints:

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

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

• 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 $x = \frac{1}{2k}$.

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Thanks. How can I reach to the 1st formula? – Michael Li Jun 8 '12 at 16:15
@Michael: I have edited my answer. – user9413 Jun 8 '12 at 16:20
It's easier to understand, thanks a lot. – Michael Li Jun 8 '12 at 16:28
@Michael: Glad that you understood. Always, welcome :) – user9413 Jun 8 '12 at 16:29

$1-\cos\left(\frac 1k\right)=\int^{1/k}_0\sin tdt$ 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.

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Integral is a little bit harsh for me to handle. Nevertheless, thanks for your help. – Michael Li Jun 8 '12 at 16:38

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 $$\lim_{x\to 0}\frac{\sin x}{x}=1$$

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