Alternating Series and P-series convergence

Click For Summary
SUMMARY

The series \(\sum^{\infty}_{n=1}\frac{-1^{n-1}}{\sqrt{n}}\) converges conditionally according to the Alternating Series Test. The term \(b(n) = \frac{1}{\sqrt{n}}\) is decreasing and approaches zero, satisfying the conditions for convergence. In contrast, the p-series test indicates that \(\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}\) diverges, confirming that the original series does not converge absolutely. Therefore, the series converges conditionally, highlighting the importance of distinguishing between absolute and conditional convergence.

PREREQUISITES
  • Understanding of the Alternating Series Test
  • Knowledge of p-series and their convergence criteria
  • Familiarity with series notation and limits
  • Basic calculus concepts, particularly limits and sequences
NEXT STEPS
  • Study the conditions for the Alternating Series Test in detail
  • Learn about absolute vs. conditional convergence in series
  • Explore examples of p-series and their convergence behavior
  • Investigate other convergence tests such as the Ratio Test and Root Test
USEFUL FOR

Students studying calculus, particularly those focusing on series convergence, as well as educators teaching convergence tests in mathematical analysis.

kylera
Messages
40
Reaction score
0
Alternating Series and P-series "convergence"

I couldn't resist trying out a pun. Anyway, onto the question:

Homework Statement


Test the series for convergence/divergence:
\sum^{\infty}_{n=1}\frac{-1^{n-1}}{\sqrt{n}}


Homework Equations


Alternating Series Test and possibly p-series test...


The Attempt at a Solution


The expression for an alternating series goes as a(n) = (-1)^(n-1) * b(n). Having said that, it's obvious that b(n) is

\frac{1}{\sqrt{n}}

which can be re-written as \frac{1}{n^\frac{1}{2}}. But by the rules of the p-series, since 0.5 is obviously lesser than 1, I came to the conclusion that the series diverges. Instead, the answer states that the series does indeed converge. Can anyone help shed some light in this?
 
Physics news on Phys.org


I think for the p-series test, the numerator must be '1' only.
 


Forget the p-series test. Concentrate on the alternating series test. b(n) is decreasing towards 0. Sorry, I don't get the 'pun'. Am I being thick?
 


Dick said:
Sorry, I don't get the 'pun'. Am I being thick?

I don't think you are since I don't get it either.
 


Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

\frac{1}{f(n)}
 


rock.freak667 said:
Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

\frac{1}{f(n)}

Sure, in fact, it applies only to series of the form 1/n^p or n^(p). I wasn't saying you were wrong. It's useful here if you want to discuss absolute convergence. It is useless for the given function.
 


The "p-series" only applies to positive series. That's why Dick says it is useful to discuss absolute convergence. If your series has both positive and negative terms then it may converge "conditionally". It will converge absolutely only if the series of absolute values converges.

By the "alternating series test", the series
\sum \frac{(-1)^n}{\sqrt{n}}
converges.

By the "p- series test"
\sum \left|\frac{(-1)^n}{\sqrt{n}}\right|= \sum \frac{1}{\sqrt{n}}= \sum \frac{1}{n^{1/2}}
does NOT converge and so the original series converges conditionally, not absolutely.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K