watchmath
  • watchmath
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
is this open for anyone to answer?
watchmath
  • watchmath
sure
anonymous
  • anonymous
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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More answers

watchmath
  • watchmath
you mean using the Taylor series? then shouldn't we have to sigma notation?
watchmath
  • watchmath
two sigma notations
anonymous
  • anonymous
oh yep i made a mistake.
anonymous
  • anonymous
hey ty for posting it up the question but whats the answer is ∑n=1∞(−1)n+1/(2n)!n2n
anonymous
  • anonymous
posted by rsvitale
watchmath
  • watchmath
no, he made a mistake. I'll post the solution if nobody answer this :).
toxicsugar22
  • toxicsugar22
hi watchmath can u help me after him
anonymous
  • anonymous
okay lol
toxicsugar22
  • toxicsugar22
watchmath can u help me
anonymous
  • anonymous
by the way do you know how to find the general term of a series
anonymous
  • anonymous
if given something like this 3,6,10,15
anonymous
  • anonymous
i mean sequence
watchmath
  • watchmath
you need to try to see the pattern and formulate it in general.
anonymous
  • anonymous
so you basically guess
watchmath
  • watchmath
No guessing is not the right word for it. It is observing some regularity :).
watchmath
  • watchmath
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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