To what value do these series converge?

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The discussion centers on the convergence of the series sin(n)/n and cos(n)/n, with participants suggesting the use of Dirichlet's test and Fourier series for evaluation. It is concluded that both series converge to zero as n approaches infinity, given that sin(n) and cos(n) oscillate between -1 and 1. One participant mentions successfully calculating the series sin(n)/n as pi/2 - 1/2 through differentiation of the Fourier cosine series. However, they express difficulty in evaluating the series cos(n)/n, proposing a method involving the evaluation of z^n/n and subsequent substitutions. The conversation highlights the complexity of integrating these series and the mathematical techniques involved.
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Consider these series:
sin(n)/n; and
cos(n)/n.
To what value do they converge? You can use Diriclet's test to show that they do converge, but to what? I think I should use a Fourier trigonometric series, but not certain how.
 
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Without ado, the answer must be "zero", because sin(n) and cos(n) can only have values between -1 to +1, while n tends to infinity.
 
I meant series, not sequence. With summation n = 1 to infinity. Sorry for any confusion.
 
Note: I'm not sure if this actually works, but it's an idea.

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\sum cos kx = -.5 + (.5sin[(2n +1)x/2)] )/sin x/2

So, if you integrate the sum, from 0 to 1, you should get

\sum (sin k)/k , which is the sum you want, so integrate the other side from 0 to 1.
 
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Thanks for the idea. I am having a hard time to find a way to integrate that expression. Meantime I was able to calculate the series sin(n)/n by differentiating the fouries cosine series cos(nx)/n^2 and substituting 1. It equates to pi/2-1
/2. I am still stuck trying to evaluate cos(n)/n however.
 
First evaluate \sum z^n/n then substitute in that to evaluate \sum e^{int}/n, then take real and imaginary parts to get \sum \cos(nt)/n and \sum \sin(nt)/n
 
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