Apr 14, 2008 · sin(n)/n already looks a lot like (-1)^n/n so I tried to figure out a way to get (-1)^n to alternate like sin(n) does. My first impulse was to use (-1)^(n/pi) but that leads to a problem …

Jul 3, 2011 · I am trying to see if a_n:={|Sin(n)|}, with n=1,2,... and | . | standard absolute value, is convergent. I know the set {k.pi}, k=1,2,... is dense in [0,1] (pi is equidistributed mod1) , and …

Feb 18, 2024 · I could summarize this as ##\frac{1}{n\pi}## for ##n=1, 5, 9, \ldots## ##\frac{-1}{n\pi}## for ##n=3, 7, 11, \ldots## and ##0## for all other ##n##.

Oct 23, 2014 · a n = [sin(n)]/n Prove that this sequence converges using Cauchy theorem Homework Equations Cauchy theorem states that: A sequence is called a Cauchy theorem if …

Mar 9, 2011 · lim (sin⁡(n)/n)=0. The instruction say I can only use the definition of limit and no additional theorems. So the first thing I should do is figure out if l sin(n)/n l < epsilon, find out …

Feb 28, 2012 · ok i know that sin n*(pi/2) = 1 if n=1,5,9,13... = -1 if n=3,7,11,15... = 0 if n is even cos n*(pi/2) = 0 if n is odd = -1 if n=0,4,8,12 = 1 if n=2,6,10,14... is there a simpler way of …

Oct 11, 2013 · How to prove that the limit \\lim_{n\\to\\infty}sin(n)^n n integer towards infinity does not exist ? If n is a real then it's obvious since we can take n=Pi/2*k k being an integer. But if n …

Apr 14, 2008 · The product of a sequence with bounded partial sums (sin(n), use a trig identity) and a function of bounded variation (1/n), converges as a series. This is proved using Abel's …

Apr 27, 2005 · This means that for any positive integer n, sin n will lie between -1 and 1. Therefore, the supremum of sin n cannot be larger than 1. Next, we can use the definition of …

Oct 12, 2008 · Yes, sin(x) is a non-monotonic function. If You intended n to be an integer, sin n is still non-monotonic because sin(0)= 0 and sin(1)= .8414... so it is increases from 0 to 1 but …