Test for Convergence for infinite Series?

Question

I'm stuck on one of my homework problems where I need to find whether the infinite series of sing(1/n)/n starting at n=1 converges or diverges. I think I have to use the ratio test since it is arbitrary but I'm not sure.

I'm talking about the infinite series of sin(1/x)/x which is the addition of all the terms resulting from whole numbers from x=1 to infinity. I don't think you can compare sin(1/x)/x to 1/x since sin(1/x)/x is arbitrary which means you cannot use the comparison test.

Answer 1

you mean sin (1/n)/n.
Compare with a(n) = 1/n^2
sin (1/n)/n /(1/n^2) = sin(1/n)/(1/n) ==>1 so the are of the same class and 1/n^2 is convergent.

Answer 2

If you mean sin(1/n)/n, then note that -1 ≤ sin(n) ≤ 1 for any n. Therefore, -1/n ≤ sin(1/n)/n ≤ 1/n, and you know that 1/n converges to zero as n → ∞.

* * * *

OK, sorry. I think the next poster has the right idea. My thinking would be to use the Taylor series sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... Using that, sin(1/x) = 1/x - 1/(3!x^3) + 1/(5!x^5) - 1/7!x^7 + ..., and sin(1/x)/x = 1/x^2 - 1/(3!x^4) + 1/(5!x^6) - 1/7!x^8 + .... This summation cannot exceed the summation of 1/x^2, and we know 1/x^2 is convergent.

Answer 3

Is there meant to be an n in the numerator? it particularly is form of a wierd finding consistent - why no longer in easy terms sqrt(sixty seven)? * * * * * definite, evaluate it to a million/?n. The sequence is divergent.