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

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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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