sin(n) does not converge - an elementary proof

[@nucreation4484]

Thank you. I appreciate you acknowledging the nuances with the problem. I was working through Apostol's and hitting my head against a wall with how to prove the sum sin(log(n)) diverges. Thanks for the video. Update: I was able to prove the sum sin(log(n)) diverges by considering a similar approach to what was done in this video. The idea is to first assume that sin(log(n)) converges to 0, and show that this assumption leads to two contradictory results. (1) that cos^2(log(n)) converges to 1, and (2) that cos^2(log(2n)) converges to 0; the latter by considering the convergence of (sin(log(4n))-sin(log(n))) using trig and log properties. The problem with my proof though is that it uses the result that every subsequence of a convergent sequence converges to the same limit which is not taught in Apostol's Calculus Vol. I (where I got the problem). However the convergent subsequence result is discussed in Apostol's book on analysis.

[@DrBarker]

I'm glad it was helpful!

[@pedrosso0]

how many cases are there that the discrete limit of the sin of a multiple of n converges? what I mean is DiscreteLim_n->infty sin(a n) when does this converge? Easy examples are when a is an integer multiple of pi/2

[@pedrosso0]

Forall a s.t. sin(a)=0 so ofc that includes pi/2 times any integer. Are there complex solutions as well?

[@toddholmes594]

Why do some source claim sin and cos diverge and some say it converges.

[@AmCanTech]

Would it be similar for sin (n*pi/2)

[@DrBarker]

sin(n*pi/2) would be a periodic sequence 0,1,0,-1,0,1,0,-1,... so again the limit wouldn't converge, but this would be easier to prove e.g. using periodicity of the sine function.

[@sumittete2804]

How can you show the series sin(n) /n is not absolutely convergent?

[@sbares]

One way is to note that among any two consecutive integers n, n+1, at least one of the numbers |sin(n)| and |sin(n+1)| must be greater than sin(0.5)=0,479... Thus |sin(2n+1)|/(2n+1) + |sin(2n+2)|/(2n+2) > 0.47/2n+2 so |sin(1)|/1 + |sin(2)|/2 + |sin(3)|/3 + |sin(4)|/4 + ... > 0.47 * (1/2 + 1/4 + 1/6 + ...) = 0.47/2 * (1 + 1/2 + 1/3 + ...) = infinity.

[@MK-tn6uv]

Squeeze theorem!

[@chopnchoopn13]

Wow great proof.

[@dalitlegreenfuzzyman]

A personal fav: sin(1)=sin(n-(n-1))=sin(n)sqrt(1-sin^2(n-1))-sin(n-1)sqrt(1-sin^2(n)). So if limsin(n) DID converge then the limit of our RHS would equal zero. Leaving us with sin(1)=0…… which is silly. Similar to what you did honestly

[@DrBarker]

Very nice!

[@ShayariWorld047]

oo my god ,,,,, so long process ......

[@DrBarker]

Hope it was worth it! I'd like to make a video proving this with 2 different subsequences at some point - it might be slightly shorter, but personally I prefer the proof in this video.

[@ShayariWorld047]

@@DrBarker ❤️🙏