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

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    Try Limit comparison to $1/k^2$2012-06-08
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    The hint given in question is to consider a p-series but I failed to link them together before asking.2012-06-08
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    Is there any resonable way to calculate its limit?2012-06-08

4 Answers 4

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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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Hints:

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

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

In reply to Michael's comment:

  • 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?2012-06-08
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    @Michael: I have edited my answer.2012-06-08
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    It's easier to understand, thanks a lot.2012-06-08
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    @Michael: Glad that you understood. Always, welcome :)2012-06-08
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$1-\cos\left(\frac 1k\right)=\int^{1/k}_0\sin tdt$ 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.2012-06-08
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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?2012-06-12
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    Maybe it's a matter of the $\sim$ notation. Is it ok now? I don't see the problem.2012-06-12
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    @anon: This is a perfectly reasonable use of the limit comparison test.2012-06-12
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    @ErickWong It appears I was confusing "limit comparison test" with "comparison test."2012-06-12
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    Thanks for your post. This method is rather more obvious than the above ones.2012-06-12
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    @Michael if you found it useful, why won't you go ahead and mark it as an answer? :-)2012-06-12
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    I've done so. ^ _ ^2012-06-12
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    ## Thx buddy ##2012-06-12