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Why does this infinite series diverge?

  • Thread starter kmacinto
  • Start date
  • #1
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Homework Statement


Why does this series diverge?


Homework Equations


[itex]\sum_{n}^{\infty }\frac{-1^{(2n+2)}}{n+1}[/itex]


The Attempt at a Solution



I must be missing a rule with the -1 sign.

My logic is that for all n, the numerator = 1 since anything to the power of 2 is even and adding the 2 doesn't change the sign either. So if the numerator is always 1 and the denominator grows without bound so by the alternating series test since, L = 0 and the series is decreasing, the series should converge... My math book and maple says different though :(
 
Last edited:

Answers and Replies

  • #2
SammyS
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Homework Statement


Why does this series diverge?


Homework Equations


[itex]\sum_{n}^{\infty }\frac{-1^{(2n+2)}}{n+1}[/itex]


The Attempt at a Solution



I must be missing a rule with the -1 sign.

My logic is that for all n, the numerator = 1 since anything to the power of 2 is even and adding the 2 doesn't change the sign either. So if the numerator is always 1 and the denominator grows without bound so by the alternating series test since, L = 0 and the series is decreasing, the series should converge... My math book and maple says different though :(
As you pointed out, it's not an alternating series.

For clarity, put parentheses around the -1. _ _ _ _ (-1)

[itex]\displaystyle \sum_{n}^{\infty }\frac{(-1)^{(2n+2)}}{n+1}[/itex]
 
  • #3
48
0

Homework Statement


Why does this series diverge?


Homework Equations


[itex]\sum_{n}^{\infty }\frac{-1^{(2n+2)}}{n+1}[/itex]


The Attempt at a Solution



I must be missing a rule with the -1 sign.

My logic is that for all n, the numerator = 1 since anything to the power of 2 is even and adding the 2 doesn't change the sign either. So if the numerator is always 1 and the denominator grows without bound so by the alternating series test since, L = 0 and the series is decreasing, the series should converge... My math book and maple says different though :(
S=Ʃ(-1)2k+2/(k+1)
Try a simpler case of the above. Try the other tests. Maybe the integral test esp. something looking like a log fn differentiated.
 
Last edited:
  • #4
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I think I have it... If the numerator is positive for all n, I can just use the direct comparison test with B sub N = 1/n which is a divergent p-series and since B sub N is less than A sub N, A sub N also diverges... right?
 
  • #5
SammyS
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That sounds good to me !
 

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