2008年4月14日 · 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 since that would give you odd/even values for the exponent alternating infinitely fast.

2011年7月3日 · 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 we have that Sin(n)=Sin(n+pi), but it seems like {|Sinn|} is dense in [0,1], so that it cannot have a limit (i.e., a unique limit point).

2024年2月18日 · 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##.

2014年10月23日 · 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 for all ε > 0, there exists N , for all n > N s.t. |x n+1 - x n | < εI do not know how to approach this proof. I would appreciate some help.

2011年3月9日 · 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 what n is greater than. I can pull the 1/n out of the absolute value, but I …

2012年2月28日 · 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 expressing this? for example simple way to express cos(n*pi)=cos(-n*pi)=(-1)^n is there a...

2013年10月11日 · 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 is a integer then sin(n) is always smaller than 1, hence the …

2008年4月14日 · 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 'summation by parts', as lurflurf intimated.

2005年4月27日 · 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 supremum, which states that the supremum of a set is the smallest upper bound of that set. In this case, the set is the set of all possible values of sin n for positive integers n.

2008年10月12日 · 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 sin(2) is .9092... and sin(3)= .1411... so it decreases from 2 to 3.