7
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When I evaluate

Limit[E^(-n)*Sum[n^k/(k!),{k,0,n}], n -> ∞]

Mathematica gives me the result

Limit[Gamma[1 + n, n]/(n Gamma[n]), n -> ∞]

Actually, the answer is exactly 1/2.

Is there any way to coax Mathematica to find the limit?

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2 Answers 2

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Taking GammaRegularized[k + 1, k] (which is equivalent to the unevaluated expression you obtained) as your function, we can consider a discretized version of your sequence, say GammaRegularized[2^k + 1, 2^k], and then give that to the (undocumented!) function SequenceLimit[], which uses the Shanks transformation:

SequenceLimit[Table[N[GammaRegularized[2^k + 1, 2^k], 25], {k, 20}]]
   0.500000000000000000

(It's not a proof, but increasing the precision and/or the number of terms of the sequence does yield consistent results.)

If you look within the Numerical Calculus package and see the definition for NLimit[], you'll see that this is essentially the procedure being used within.

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  • $\begingroup$ how come Spawn1701D's answer is so inaccurate? $\endgroup$
    – chris
    Apr 14, 2013 at 13:54
  • $\begingroup$ @chris, I suppose option tweaking would be in order... $\endgroup$ Apr 14, 2013 at 13:55
  • $\begingroup$ NLimit[Gamma[1 + n, n]/Gamma[n + 1], n -> Infinity, Terms -> 8] yields 0.50007 $\endgroup$
    – chris
    Apr 14, 2013 at 13:58
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Try with NLimit:

Needs["NumericalCalculus`"]

NLimit[Gamma[1 + n, n]/Gamma[n + 1], n -> Infinity]

0.499858

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