Strange Sequence

  • #1
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Homework Statement


Find sequence [tex](a_n)[/tex] s.t. [tex]\lim_{N\rightarrow \infty} \sum_{n=1}^{2N} a_n[/tex] and [tex]\lim_{N\rightarrow \infty} \sum_{n=1}^{2N+1} a_n[/tex] both converge but [tex]\sum_{n=1}^{\infty} a_n[/tex] diverges.




I have no idea where to start to be honest. I'm confused at how this is possible. Isn't it always just summed to infinity regardless of whether it's 2(x) or 2(x+1) where x-> infinity?
 

Answers and Replies

  • #2
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[tex]a_n=(-1)^n[/tex]
 
  • #3
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Can you explain how that works? I'd rather know what the problems means/is asking instead of really knowing the answer. I was guessing it had to do with alternating series, but I don't know what differentiates 2N, 2N + 1, and N.

Or does it mean that:
for the first sequence
(-1)^n = -1 + 1 + -1 + ... + (-1)^(2N) = (-1)^(2N) -> 1 as n approaches infinity
and same logic for the rest?
That would make sense I guess.

Edit: actually since there's an even number of terms for the first sequence and it goes -1,1, etc., then it should converge to 0. I would assume this is the correct logic to follow, but how is shown rigorously? Clearly, there must be something wrong in saying there is an even # of infinite terms.
 
Last edited:
  • #4
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[tex]\sum_{n=1}^{N}a_n=(-1)^N[/tex]
so the series doesn't converge. On the other hand
[tex]\sum_{n=1}^{2N}a_n=0\qquad\sum_{n=1}^{2N+1}a_n=-1[/tex]
so both limits exist. Sorry I left it w/o explaination, I though it was this kind of problems where the answer is kinda tricky, but easy to verify.
 
  • #5
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Yeah, after a bit longer, I recognized the logic. Thanks a bunch! Is there a way to rigorously explain that [tex]\lim_{N\rightarrow \infty} \sum_{n=1}^{2N} a_n=0 [/tex]?
 
  • #6
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Since each member of this sequence is zero, the limit is zero as well.
 
  • #7
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I'm not really sure how you define member since in my belief, each individual term is -1 or 1. However, I'm assuming you mean member is not the same as term.
Is it valid to say [tex]
\lim_{N\rightarrow \infty} \sum_{n=1}^{2N} a_n= (-1) + 1 + (-1) + 1 + ... = (-1 + 1) + (-1 + 1) + ... = \sum_{n=1}^{\infty} (-1+1) = \sum_{n=1}^{\infty} 0 = 0
[/tex]?
 
  • #8
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[tex]\sum_{n=1}^{2N}a_n=0[/tex] for every [tex]N[/tex], and so,
[tex]\lim_{N\rightarrow\infty}\sum_{n=1}^{2N}a_n=\lim_{N\rightarrow\infty}0=0[/tex]
 
  • #9
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Oh. Makes sense. Thanks.
 

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