Absolute convergence help

  • Thread starter miglo
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  • #1
miglo
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Homework Statement


[tex]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sqrt{n}+1}{n+1}[/tex]


Homework Equations


absolute convergence test


The Attempt at a Solution


by book says that the series converges because [tex]\sum_{n=1}^{\infty}\frac{\sqrt{n}+1}{n+1}[/tex] converges
but they don't show how the absolute value of the original series converges, and I've tried showing it myself but i keep getting divergence
i know that as n grows larger and larger the behavior of [tex]\frac{\sqrt{n}+1}{n+1}[/tex] is similar to that of [tex]\frac{\sqrt{n}}{n}[/tex] so i tried using limit comparison and direct comparison with [itex]\frac{1}{n}[/itex] but i keep getting divergence
i tried the integral test but i kept getting divergence also
ive been trying this for far too long so any help would be greatly appreciated
 

Answers and Replies

  • #2
LCKurtz
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You are correct that the positive term series diverges.
 
  • #3
miglo
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but my book says that the original series converges by the absolute convergence test
so wouldn't that mean that [tex]\sum_{n=1}^{\infty}\frac{\sqrt{n}+1}{n+1}[/tex] converges also? or is this an error in the book?
 
  • #4
LCKurtz
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but my book says that the original series converges by the absolute convergence test
so wouldn't that mean that [tex]\sum_{n=1}^{\infty}\frac{\sqrt{n}+1}{n+1}[/tex] converges also? or is this an error in the book?

The series is not absolutely convergent. It may be convergent with the alternating signs in which case it would be called "conditionally convergent". (I didn't check that). But the positive term series you are asking about is definitely divergent. You know it is because you correctly checked it.
 
  • #5
miglo
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well then ill just check to see if it convergences by the alternating series test
thanks a lot!
 

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