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Re: Re: Boundary cond. at Infinity



I believethat this notebook is a clearer and more accurate response than
my first response.

Bob Hanlon
__________________________

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  ") is negative, the terms of the hypergeometric series alternate
signs.  \ This is not a convenient form for determining the limit. 
Using a linear \ transformation (Abramowitz and Stegun, 15.3.5) to
obtain a positive \ argument:"
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And the denominator tends to Infinity.  For a>1 the limit as x goes \ to
+Infinity is then zero.\
\>", "Text"],

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