Does $$\sum_{k=1}^{\infty} \left(1-\cos\frac{1}{k}\right)$$ converge or diverge?
Convergence or divergence of $\sum_{k=1}^{\infty} \left(1-\cos\frac{1}{k}\right)$
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3Try Limit comparison to $1/k^2$ – 2012-06-08
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0The 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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0Is there any resonable way to calculate its limit? – 2012-06-08
4 Answers
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 $$\lim_{x\to 0}\frac{\sin x}{x}=1$$
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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0Thanks. How can I reach to the 1st formula? – 2012-06-08
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0@Michael: I have edited my answer. – 2012-06-08
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0It's easier to understand, thanks a lot. – 2012-06-08
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0@Michael: Glad that you understood. Always, welcome :) – 2012-06-08
$1-\cos\left(\frac 1k\right)=\int^{1/k}_0\sin tdt$ 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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0Integral is a little bit harsh for me to handle. Nevertheless, thanks for your help. – 2012-06-08
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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0$\frac{1}{2}f+o(f)$ is not $\sim f$. And are you sure that's the comparison test? – 2012-06-12
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0Maybe 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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0@anon: This is a perfectly reasonable use of the limit comparison test. – 2012-06-12
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0@ErickWong It appears I was confusing "limit comparison test" with "comparison test." – 2012-06-12
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0Thanks for your post. This method is rather more obvious than the above ones. – 2012-06-12
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0@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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0I've done so. ^ _ ^ – 2012-06-12
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0## Thx buddy ## – 2012-06-12