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

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this open for anyone to answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0you mean using the Taylor series? then shouldn't we have to sigma notation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yep i made a mistake.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey ty for posting it up the question but whats the answer is ∑n=1∞(−1)n+1/(2n)!n2n

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0no, he made a mistake. I'll post the solution if nobody answer this :).

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0hi watchmath can u help me after him

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath can u help me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by the way do you know how to find the general term of a series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if given something like this 3,6,10,15

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0you need to try to see the pattern and formulate it in general.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you basically guess

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0No guessing is not the right word for it. It is observing some regularity :).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 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 (pseries with p=2). Hence \(\sum 2\sin^2(1/(2n))\) is convergent as well :).
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