Does the Series Converge or Diverge for Different Values of z?

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SUMMARY

The series $$\sum _{p=1}^{\infty }\dfrac {z^{p}} {p^{3/2}}$$ converges for values of z equal to 1 and less than 1, while it diverges for values greater than 1. The ratio test is the appropriate method to determine convergence or divergence, specifically by evaluating the limit $$\lim_{p \to \infty} \frac{a_{p+1}}{a_{p}}$$. Misapplication of the ratio test can lead to incorrect conclusions about the behavior of the series.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the Ratio Test in calculus
  • Knowledge of limits and their evaluation
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review the Ratio Test for series convergence
  • Study examples of series that converge and diverge based on their terms
  • Learn about other convergence tests such as the Root Test and Comparison Test
  • Explore the implications of series convergence in real-world applications
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Students studying calculus, mathematicians analyzing series, and educators teaching convergence concepts will benefit from this discussion.

erbilsilik
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Homework Statement



How can I show that this series is convergent for z=1 and z<1 and divergent for z>1

$$\sum _{p=1}^{\infty }\dfrac {z^{p}} {p^{3/2}}$$

Homework Equations



http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

The Attempt at a Solution



Using the ratio test I've found:

$$\lim _{p\rightarrow \infty }\sum _{p=1}^{\infty }\dfrac {z^{p}} {\left( p+1\right) ^{3/2}}$$
[/B]
 
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You applied the ratio test wrongly. Given a series
\sum_{p = 1}^{\infty} a_{p},
the ratio test involves looking at the quantity
\lim_{p \to \infty} \frac{a_{p+1}}{a_{p}}.

If this quantity is greater than one, then the series diverges.
 

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