- #1
icesalmon
- 270
- 13
Homework Statement
given an = ( -1 )(n+1) / [itex]\sqrt{n}[/itex] determine if the infinite series is Absolutely Convergent, Conditionally Convergent, or Divergent.
Homework Equations
I hope I have these theorems down correctly, please correct me if I'm wrong. If [itex]\Sigma[/itex]|an| is Convergent then [itex]\Sigma[/itex]an is Absolutely Convergent if [itex]\Sigma[/itex]|an| is Divergent and [itex]\Sigma[/itex]an is Convergent then [itex]\Sigma[/itex]an is Conditionally Convergent And if neither occur then the [itex]\Sigma[/itex]an may be Divergent.
The Attempt at a Solution
To show [itex]\Sigma[/itex]|an| or [itex]\Sigma[/itex]an are convergent I have to show 2 things, 1). lim[ n[itex]\rightarrow[/itex][itex]\infty[/itex] ] |an| = 0 and that either an+1 [itex]\leq[/itex] an, an+1/an [itex]\leq[/itex] 1 or d(an)/dn < 0 which determine whether or not an is Decreasing. If I can show these things I have a convergent Alternating Series, if not I might be able to try another test or I have a divergent Alternating Series.
So I started by seeing if lim[ n [itex]\rightarrow[/itex] [itex]\infty[/itex] ] |an| = 0 so I have to figure out what |an| actually is. Again correct me if I'm wrong, it would help me out greatly here, but I believe |(-1)n+1/[itex]\sqrt{n}[/itex]| to be 1/[itex]\sqrt{n}[/itex] because |+/-(1)| = 1 there for I can determine that lim[ n [itex]\rightarrow[/itex] [itex]\infty[/itex] ] 1/[itex]\sqrt{n}[/itex] = 1/[itex]\infty[/itex] = 0 so if I have my thoughts in order I have shown lim[n [itex]\rightarrow[/itex] [itex]\infty[/itex] ] 1/[itex]\sqrt{n}[/itex] = 0 great! Next I have to show that an+1 [itex]\leq[/itex] an so I first considered [itex]\sqrt{n+1}[/itex] which[itex]\geq[/itex] [itex]\sqrt{n}[/itex] so 1/|[itex]\sqrt{n+1}[/itex]| [itex]\leq[/itex] 1/|[itex]\sqrt{n}[/itex]|
to further solidify this in my mind or at least on paper, I wanted to show that d(|an|/dn [itex]\leq[/itex] 0 so I did just that and acquired -1/(2n3/2) [itex]\leq[/itex] 0 [itex]\forall[/itex] [itex]\Re[/itex] therefor [itex]\Sigma[/itex]an is Absolutely Convergent. The book says that it is conditionally convergent, I tried the integral test for no good reason and paired with my above results got the exact opposite of my definition for conditional convergence.
After moving on I got the subsequent problems incorrect as well, falsly concluding that my series was Absolutely Convergent when they were Conditionally Convergent so I can't really move anywhere until I understand this, thanks in advance for your help.