Is this convergent? sum (sin k)

In summary, the conversation discusses the convergence of the series \sum sin k and k is from 1 to infinity, and how it is not necessarily divergent. The concept of "diverge" is also debated, with some considering it to mean "diverges to infinity" and others considering it to simply mean "does not converge". The conversation also brings up the example of the series \sum\sin(\pi k), which does converge, and how the sum becomes 0 at every pi. However, the concept of infinity and its moduli are also discussed, leading to the conclusion that the series is oscillating.
  • #1
pliu123123
43
0
[itex]\sum sin k[/itex]
and k is from 1 to infinity,thx
 
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  • #2
What do you think?? What tests for series can you apply to this?
 
  • #3
pliu123123 said:
[itex]\sum sin k[/itex]
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.
 
  • #4
dimension10 said:
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.
 
  • #5
Mute said:
A sum that doesn't converge doesn't necessarily diverge.
Erm, are you sure?
 
  • #6
I could have sworn that the definition of "diverge" was "doen't converge"!

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

has partial sums, 1, 0, 1, 0, 1, 0, ... which does not "diverge to infinity" but does "diverge".
 
  • #7
HallsofIvy said:
I could have sworn that the definition of "diverge" was "doen't converge"!

Which does NOT necessarily mean "diverge to infinity".
[tex]\sum_{n=0}^\infty (-1)^n[/tex]= 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 [itex]\pm \infty[/itex]" and something with partial sums which have no limit simply had no limit.
 
  • #8
Mute said:
I stand corrected, then! I've always considered "diverge" to be short for "diverges to [itex]\pm \infty[/itex]" 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.
 
  • #9
Ok, it is an oscillating series, at least.
 
  • #10
dimension10 said:
Ok, it is an oscillating series, at least.

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

[tex]\sum\sin(\pi k)[/tex]

does converge. I always ask this as exam question, only the ones who know what they're doing answer this correctly.
 
  • #11
micromass said:
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?

micromass said:
For example, the series

[tex]\sum\sin(\pi k)[/tex]

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)
 
  • #12
dimension10 said:
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 :frown:

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

Yes.
 
  • #13
micromass said:
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 :frown:

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.
 

Related to Is this convergent? sum (sin k)

1. What is the definition of convergence in mathematics?

The concept of convergence in mathematics refers to the idea that a sequence or series of numbers approaches a definite value or limit as more terms are added. In other words, as the terms of the sequence or series get closer and closer together, they eventually reach a fixed value.

2. How can I tell if a series is convergent or divergent?

There are several methods for determining the convergence or divergence of a series, including the ratio test, integral test, and comparison test. These methods involve comparing the series to known convergent or divergent series or using mathematical calculations to determine the behavior of the series.

3. Is the series sum (sin k) convergent?

The series sum (sin k) is known as the sine series and it is a divergent series. This can be shown using the limit comparison test or by graphing the series and observing that it does not approach a definite value as more terms are added.

4. Are there any special cases where a divergent series can be considered convergent?

Yes, there are some cases where a divergent series can be considered convergent. One example is the Grandi's series, which is the series 1 - 1 + 1 - 1 + 1 - 1 + ... This series does not have a definite value, but it can be assigned a value of 1/2 using a mathematical technique called Cesàro summation.

5. How is the convergence of series related to real-world applications?

The concept of convergence is widely used in various fields of science and engineering, including physics, economics, and computer science. In physics, the convergence of series is used to calculate quantities such as electric potential and magnetic field. In economics, it is used to determine the value of investments over time. In computer science, the convergence of series is used in algorithms for numerical analysis and data compression.

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