Determine the convergence of the series \[\sum_{n=1}^\infty 1-\cos(1/n)\] (this is a problem from one of the student here)

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Determine the convergence of the series \[\sum_{n=1}^\infty 1-\cos(1/n)\] (this is a problem from one of the student here)

Mathematics
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i expanded cos(1/n) and rewrote the summand as: \[\sum_{n=1}^{\infty}(-1)^{n+1}/(2n)!n^{2n}\] which is an alternating series that is monotone decreasing, so it converges.

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Other answers:

you mean using the Taylor series? then shouldn't we have to sigma notation?
two sigma notations
oh yep i made a mistake.
hey ty for posting it up the question but whats the answer is ∑n=1∞(−1)n+1/(2n)!n2n
posted by rsvitale
no, he made a mistake. I'll post the solution if nobody answer this :).
hi watchmath can u help me after him
okay lol
watchmath can u help me
by the way do you know how to find the general term of a series
if given something like this 3,6,10,15
i mean sequence
you need to try to see the pattern and formulate it in general.
so you basically guess
No guessing is not the right word for it. It is observing some regularity :).
Ok, let solve the series problem. By the double angle formula we know that 1- cos(2x)= 2sin^2(x) It follows that \(1-\cos(1/n)=2\sin^2(1/(2n))\) Recall that for positive x we have \(\sin x < x\) It follows that \(\sin^2(1/(2n))< (1/(2n) )^2\) But know the series \(\sum 1/(2n)^2\) is convergent (p-series with p=2). Hence \(\sum 2\sin^2(1/(2n))\) is convergent as well :).

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