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Does (-1)^n Converge?

  • Thread starter zmike
  • Start date
  • #1
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Intuitively I would think that [SUM of (-1)^n] or [SIGMA (-1)^n as n->infinity] Converges to 0 but apparently it doesn't according to the test for divergence ????

Does it actually converge or diverge??? Why?

Thanks
 

Answers and Replies

  • #2
dx
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Hi zmike,

It doesn't converge:

1 = 1
1 + (-1) = 0
1 + (-1) + 1 = 1
1 + (-1) + 1 + (-1) = 0

It keeps oscillating between 1 and 0.
 
Last edited:
  • #3
102
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Hi zmike,

It doesn't converge:

1 = 1
1 + (-1) = 0
1 + (-1) + 1 = 1
1 + (-1) + 1 + (-1) = 0

It keeps oscillating between 1 and 0.
thanks, just one more thing,

so for ratio test, root test, test for divergence ALL work for alternating series? I am asking this b/c my textbook says they only work for positive sequences

also can I seperate a sum? so can I take
[sigma an] * [sigma bn] = sigma (an*bn)? If not, wouldn't that violate the properties of limits?

Thanks
 
  • #4
dx
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The ratio test applies only to positive series.

For finite sums (∑a)(∑b) = ∑∑(ab) is true.
 
  • #5
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(∑a)(∑b) = ∑∑(ab)

What does the ∑∑ mean? take the sum twice? so is this also true with infinite sums?

also, does this mean that if either a or b is divergent, I can conclude that the series is divergent?

thanks
 
  • #6
dx
Homework Helper
Gold Member
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Sorry about the confusing notation. Latex is not working currently and I didn't know how to write it.

Sum(i = 1 to n)[A_i] x Sum(j = 1 to m)[B_j] equals Sum(i = 1 to n)Sum(j = 1 to m)[A_i B_j]

For infinite series, if both the series are convergent, this will still work. If either of them is divergent, then clearly the left hand side is not defined.
 
Last edited:

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