Absolute or Conditional Convergence, or Divergence of Alternating Series.

[icesalmon]

Homework Statement

given a_n = ( -1 )⁽ⁿ⁺¹⁾ / \sqrt{n} 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 \Sigma|a_n| is Convergent then \Sigmaa_n is Absolutely Convergent if \Sigma|a_n| is Divergent and \Sigmaa_n is Convergent then \Sigmaa_n is Conditionally Convergent And if neither occur then the \Sigmaa_n may be Divergent.

The Attempt at a Solution

To show \Sigma|a_n| or \Sigmaa_n are convergent I have to show 2 things, 1). lim[ n\rightarrow\infty ] |a_n| = 0 and that either a_n+1 \leq a_n, a_n+1/a_n \leq 1 or d(a_n)/dn < 0 which determine whether or not a_n 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 \rightarrow \infty ] |a_n| = 0 so I have to figure out what |a_n| actually is. Again correct me if I'm wrong, it would help me out greatly here, but I believe |(-1)ⁿ⁺¹/\sqrt{n}| to be 1/\sqrt{n} because |+/-(1)| = 1 there for I can determine that lim[ n \rightarrow \infty ] 1/\sqrt{n} = 1/\infty = 0 so if I have my thoughts in order I have shown lim[n \rightarrow \infty ] 1/\sqrt{n} = 0 great! Next I have to show that a_n+1 \leq a_n so I first considered \sqrt{n+1} which\geq \sqrt{n} so 1/|\sqrt{n+1}| \leq 1/|\sqrt{n}|
to further solidify this in my mind or at least on paper, I wanted to show that d(|a_n|/dn \leq 0 so I did just that and acquired -1/(2n^3/2) \leq 0 \forall \Re therefor \Sigmaa_n 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.

 

[lanedance]

i think you're on the right track, the series is conditionally convergent shown by applying the "alternating series test"
-an tends to zero monotonically and changes sign
so the series as its written it converges to a finite limit

however as you've also correctly worked out it is not absolutely convergent as the sum of the modulus of each term diverges

any absolutely convergent series is conditionally convergent, however the converse does not apply