Another infinite series

  • #1
I am having a difficult time working with some of these infinite series. I studied them in calc 2, but that was a few years ago.

Could someone help me figure out how to find what the following sum converges to:

[tex]
\sum_{n=-N}^N{cos(n \theta)}
[/tex]

Shouldn't there be some property or convergence test I could do to find out what it actually converges to?
 
Last edited:

Answers and Replies

  • #2
This expands to:

[tex]
cos(-N\theta)+ ... +cos(-4\theta)+ cos(-3\theta)+ cos(-2\theta) + cos(-\theta) + cos(0)
+ cos(\theta) + cos(2\theta) + cos(3\theta)+ cos(4\theta)+ ... + cos(N \theta)
[/tex]


And since [tex] cos(-\alpha) = cos(\alpha) [/tex] the above expansion boils down to:

[tex]
cos(N\theta)+ ... +cos(4\theta)+ cos(3\theta)+ cos(2\theta) + cos(\theta) + cos(0)
+ cos(\theta) + cos(2\theta) + cos(3\theta)+ cos(4\theta)+ ... + cos(N \theta)
[/tex]

Which can be simplified as:

[tex]
cos(0) + 2cos(\theta) + 2cos(2\theta) + 2cos(3\theta)+ 2cos(4\theta)+ ... + 2cos(N \theta)
[/tex]
 
Last edited:
  • #3
So I am actually looking at the case where [tex] \theta = \frac{\pi}{2} [/tex]

Therefore:

[tex]
2cos(0) = 2
[/tex]

[tex]
2cos( \frac{\pi}{2} ) = 0
[/tex]

[tex]
2cos(\pi) = -2
[/tex]

[tex]
2cos( \frac{3\pi}{2} ) = 0
[/tex]

[tex]
2cos( 2\pi ) = 2
[/tex]

[tex] . [/tex]

[tex] . [/tex]

[tex] . [/tex]

[tex]
2cos(N \frac{\pi}{2})
[/tex]

So the sequence would be:
[tex]
\{2, 0, -2, 0, 2, ... , 2cos(N \frac{\pi}{2}) \}
[/tex]

Is my logic correct?
And if so, how do I sum the sequence to find what it converges to?
 
Last edited:
  • #4
mathman
Science Advisor
7,869
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So I am actually looking at the case where [tex] \theta = \frac{\pi}{2} [/tex]

Therefore:

[tex]
2cos(0) = 2
[/tex]

[tex]
2cos( \frac{\pi}{2} ) = 0
[/tex]

[tex]
2cos(\pi) = -2
[/tex]

[tex]
2cos( \frac{3\pi}{2} ) = 0
[/tex]

[tex]
2cos( 2\pi ) = 2
[/tex]

[tex] . [/tex]

[tex] . [/tex]

[tex] . [/tex]

[tex]
2cos(N \frac{\pi}{2})
[/tex]

So the sequence would be:
[tex]
\{2, 0, -2, 0, 2, ... , 2cos(N \frac{\pi}{2}) \}
[/tex]

Is my logic correct?
And if so, how do I sum the sequence to find what it converges to?
Your logic is correct. The conclusion is that it doesn't converge.
 
  • #5
Okay. Well this is part of a homework problem, so if an admin could move this to the homework section in "calculus and beyond" that would be great.

Here is the whole problem:

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{|cos(n \frac{\pi}{4})|^2}
[/tex]


Here is how far I have evaluated it:

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{cos^2(n \frac{\pi}{4})}
[/tex]

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{\frac{1+cos(2n \frac{\pi}{4})}{2}}
[/tex]

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
(\sum_{n=-N}^N{\frac{1}{2}} +
\sum_{n=-N}^N{\frac{1}{2}cos(n \frac{\pi}{2})}})
[/tex]

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{\frac{1}{2}} +
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{\frac{1}{2}cos(n \frac{\pi}{2})}}
[/tex]

[tex]
\frac{1}{2}
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}(2N+1)+
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
(\frac{1}{2})
\sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
[/tex]

[tex]
\frac{1}{2}
\lim_{N \rightarrow \infty}(1)+
\frac{1}{2}
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
[/tex]

[tex]
\frac{1}{2}+
\frac{1}{2}
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
[/tex]
 
Last edited:
  • #6
The final answer is
[tex]
\frac{1}{2}
[/tex]

So as long as I have done everything else correct so far this part should go to zero:

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
[/tex]

Is that correct?
And if so, how does that go to zero?
 
  • #7
Maybe that is as far as I can simplify:

[tex]
\sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
[/tex]

So I should evaluate the limit on the outside. That would make:
[tex]

\lim_{N \rightarrow \infty}
\frac{1}{2N+1}=0
[/tex]


Therefore making the entire term:

[tex]
\lim_{N \rightarrow \infty}
\frac{1}{2N+1}
\sum_{n=-N}^N{cos(n \frac{\pi}{2})}} = 0
[/tex]

Is that correct logic?
 
  • #8
mathman
Science Advisor
7,869
450
I suggest you go back to start.
cos2(nπ/4) =1 for n=0, 1/2 for n=1, 0 for n=2, 1/2 for n=3, and cycles from then on. For each set of 4 terms, the sum is 1, so the limit of the original sum is 1/4.
 
  • #9
Could you please elaborate. I don't understand.
 
  • #10
mathman
Science Advisor
7,869
450
Could you please elaborate. I don't understand.
I'm not sure how to elaborate, except to say that I made an error in arithmetic. The sum of four terms (1 + 1/2 +0 + 1/2) is 2 (not 1), so the limit = 1/2. This agrees with the result you got.
 
  • #11
64
0
Actually, you can find a close form for your summation expression. Please refer to http://www.voofie.com/concept/Mathematics/" [Broken] for details:

http://www.voofie.com/content/156/how-to-sum-sinn-q-and-cosn-q/" [Broken]

In short,

[tex]\sum _{n=-N}^N \cos (n \theta )=\cos (N \theta )+\cot \left(\frac{\theta }{2}\right) \sin (N \theta )[/tex]

With this close form, you can solve your limit problem easily.
 
Last edited by a moderator:

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