What is the sum of sin 1/k when k is approaching infinity?

Question

Is it convergent, divergent and what test if any did you use to get it.

Answer 1

I believe it is divergent. Here is my reasoning:

The power series of sin(1/k) is:
1/k - (1/3!)(1/k^3) + (1/5!)(1/k^5) - ...

Notice that the sum of each term except the first is convergentm but the sum of the first terms is divergent. So the sum of sin(1/k) is divergent.

Edit: Very neat picture modulo_function, what is it from? Perhaps I should have been a little bit more specific. Although the power series shows that sinx < 1/k, what we can say is that the power series at least 1/k - 1/k^3. Then since the sum of 1/k diverges and the sum of 1/k^3 converges, you basically have infinity minus some constant, which is infinity, so the sum of (1/k - 1/k^3) diverges, and so does the sum of sin(1/k).

Answer 2

Ben is correct.
Since Σ(1/k) does not converge, Σsin(1/k) cannot converge.

This can also be demonstrated by observing that
(sin(1/k))/sin(1/(k - 1)) < (sin(1/(k + 1))/sin(1/k) for any k > 1, so the series fails the test for convergence.

Answer 3

I'm not going along with Ben on this one.

For small x:
sin x <= x

which means that:
For large k:

sin 1/k < 1/k therefore the harmonic series cannot be used to show divergence.

However, my intuition agrees with Ben.

I remember doing a similar problem in my real analysis class, which I need to review for a current class, and so I'll see what I can find.