Is this convergent? sum (sin k)

[member 141513]

\sum sin k
and k is from 1 to infinity,thx

 

[micromass]

What do you think?? What tests for series can you apply to this?

 

[dimension10]

\sum sin k
and k is from 1 to infinity,thx

Nope. Its divergent. From 1 to 2pi, it is x. For 1 to infinty it is infinity x=infinity.

 

[Mute]

Nope. Its divergent. From 1 to 2pi, it is x. For 1 to infinty it is infinity x=infinity.

Are you sure it diverges? A sum that doesn't converge doesn't necessarily diverge.

 

[Hootenanny]

A sum that doesn't converge doesn't necessarily diverge.

Erm, are you sure?

 

[HallsofIvy]

I could have sworn that the definition of "diverge" was "doen't converge"!

Which does NOT necessarily mean "diverge to infinity".
\sum_{n=0}^\infty (-1)^n= 1- 1+ 1- 1+ 1- 1+...

has partial sums, 1, 0, 1, 0, 1, 0, ... which does not "diverge to infinity" but does "diverge".

 

[Mute]

I could have sworn that the definition of "diverge" was "doen't converge"!

Which does NOT necessarily mean "diverge to infinity".
\sum_{n=0}^\infty (-1)^n= 1- 1+ 1- 1+ 1- 1+...

has partial sums, 1, 0, 1, 0, 1, 0, ... which does not "diverge to infinity" but does "diverge".

I stand corrected, then! I've always considered "diverge" to be short for "diverges to \pm \infty" and something with partial sums which have no limit simply had no limit.

 

[gb7nash]

I stand corrected, then! I've always considered "diverge" to be short for "diverges to \pm \infty" and something with partial sums which have no limit simply had no limit.

And to throw another monkey wrench into this, if we're working with the extended real numbers, and the limit went to infinity, then it would converge to infinity. Similarly for -infinity.

 

[dimension10]

Ok, it is an oscillating series, at least.

 

[micromass]

Ok, it is an oscillating series, at least.

Yes it is, but proving that might not be so obvious.
For example, the series

\sum\sin(\pi k)

does converge. I always ask this as exam question, only the ones who know what they're doing answer this correctly.

 

[dimension10]

Yes it is, but proving that might not be so obvious.

At every pi, the sum becomes 0. infinity+pi/2=infinity. So infinity could just be a multiple of 2pi but it could have a modulus of pi/2. sin(pi/2)=1. So is this why the series is oscillating?

For example, the series

\sum\sin(\pi k)

does converge. I always ask this as exam question, only the ones who know what they're doing answer this correctly.

Does it converge to 0? because sin(pi*k)=0 (when k is an integer)

 

[micromass]

At every pi, the sum becomes 0. infinity+pi/2=infinity. So infinity could just be a multiple of 2pi but it could have a modulus of pi/2. sin(pi/2)=1. So is this why the series is oscillating?

Huh? First of all, you take sin(k) for integers k, so k never becomes pi (or an integer multiple of pi). So the sum never becomes zero. The rest of your post doesn't make much sense to me

Does it converge to 0? because sin(pi*k)=0 (when k is an integer)

Yes.

 

[dimension10]

Huh? First of all, you take sin(k) for integers k, so k never becomes pi (or an integer multiple of pi). So the sum never becomes zero. The rest of your post doesn't make much sense to me

Ok, at least there are multiple moduli for infinity when divided by an integer thus the series is oscillating. Still using the same method though.