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Why is the output of the limit below an interval? It should be precisely $1$.

Limit[(2/Pi) ((2 n + 2)!!/(2 n + 1)!!) Integrate[(1 - x^2)^(n + 1/2), {x, 0, 1}],
   n -> Infinity]
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1 Answer 1

up vote 7 down vote accepted

You have made an assumption without telling Mathematica about it. Namely, that $n$ is an integer. You can add this assumption like so:

Limit[(2 (2n + 2)!! Integrate[(1 - x^2)^(n + 1/2), 
          {x, 0, 1}])/(Pi(2n + 1)!!), n -> Infinity, 
   Assumptions -> n ∈ Integers]
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thank you for your explanations! –  Chris's sis Feb 27 '13 at 23:39
1  
Without this assumption it appears that the correct interval of fluctuation is 2/Pi to 1. Limit[] is unable to completely sort out the dependencies of the various fluctuating components. I think it comes reasonably close though; if you blindly plug in for n an interval of length 2Pi, for large n, the interval range that results is far bigger. –  Daniel Lichtblau Feb 28 '13 at 15:06
    
@Daniel Using FunctionExpand on the expression first gives {2/Pi, 1}. Does FunctionExpand do any transformations that are not generally valid? –  Szabolcs Mar 1 '13 at 0:35
    
@Szabolcs I don't think it does anything invalid. I had used FunctionExpand myself, but maybe I'm not using it in the same way or place you did. –  Daniel Lichtblau Mar 1 '13 at 15:27
    
@Danel I used it as Limit[FunctionExpand[(2/ Pi) ((2 n + 2)!!/(2 n + 1)!!) Integrate[(1 - x^2)^(n + 1/2), {x, 0, 1}]], n -> \[Infinity]] I was just wondering why Limit doesn't try to use it itself. But how these symbolic functions can choose a strategy to calculate something is a big mystery to me :) It must be quite complicated and difficult to get right. –  Szabolcs Mar 1 '13 at 15:51

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