Convergence of Infinite Series

AI Thread Summary
The discussion revolves around the convergence of an infinite series defined by a specific formula. The user applies D'Alembert's ratio test, concluding that the series diverges for x > 1 and converges for x < 1. When x = 1, they utilize Raabe's test, determining that the series converges if γ - α - β > 1 and diverges if γ - α - β < 1. However, the textbook states that convergence occurs if γ - α - β > 0 and divergence if γ - α - β < 0. The user seeks clarification on their apparent mistake in applying Raabe's test.
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Homework Statement


1+ \frac{\alpha\beta}{\gamma} x + \frac{\alpha (\alpha+1)\beta(\beta+1)}{1.2.\gamma(\gamma+1)}x^{2}+...


Homework Equations





The Attempt at a Solution


Using D'Alembert's ratio test, I get lim_{n\rightarrow\infty}\frac{U_{n+1}}{U_{n}}=x
so, x>1 diverging series
x<1 converging series
when x=1, Using Raabe's test I get
lim_{n\rightarrow\infty}n[\frac{U_{n}}{U_{n+1}}-1]=\gamma-\alpha-\beta
so, Series Converges if \gamma-\alpha-\beta>1
and diverges if \gamma-\alpha-\beta<1

However the book has given the answer to be
coverges if \gamma-\alpha-\beta>0
and diverges if \gamma-\alpha-\beta<0

Can anyone point out my mistake, please?
 
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For Raabe's test wtih x=1, I get
\lim_{n\rightarrow\infty}n\!\left({\textstyle{U_{n}\over U_{n+1}}-1\right)=\gamma-\alpha-\beta+1
 
My expression for U_{n} = \frac{1.\alpha...(\alpha + n-1) 1.\beta...(\beta+n-1)}{1.2...(n-1) 1.\gamma...(\gamma+n-1)}

Using this I get \gamma-\alpha-\beta from Raabe's test
 
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