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Can this expression be simplified?

$ \sum_{i=0}^n (k^i)/i! $

EDIT: This expression I have got as the number of possible ways to select n objects or less from k different infinite objects (you can select as many as you can from any type)...

I believe it must be equal to the number of possible ways to select n objects from k+1 different infinite objects, where the extra +1 is a dummy object, selecting one of this type means letting an empty space in the final selection (i.e. selecting n-1 not n)...

so this must be equal to:

$ (k+1)^n/n! $

Why that is not right??

  • 0
    Note that 1 plus this expression would be a truncation of the power series for exp(k). Perhaps you meant to start the summation at i=0? In any case this is a polynomial in k. There are more efficient ways to evaluate it than term-by-term, e.g. Horner's method. Is that the kind of "simplification" you have in mind?2010-11-30
  • 2
    That truncation of the exponential function can be expressed as an incomplete gamma, but my opinion is that's the simplest you can get for that sum, computationally speaking.2010-11-30
  • 5
    [Speaking of which...](http://math.stackexchange.com/questions/1283)2010-11-30
  • 0
    Not so relevant, but for $k=n$ this expression is asymptotically $e^n / 2$ as $n \to \infty$ (by the central limit theorem).2010-11-30
  • 0
    This is an expression not an equation.2010-11-30
  • 0
    As J. M. said, this is a possible duplicate of [Closed form of a partial sum of the power series of exp(x)](http://math.stackexchange.com/questions/1283/closed-form-of-a-partial-sum-of-the-power-series-of-expx)2010-11-30
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    Please take a look on the edited question..2010-11-30
  • 0
    "k infinite objects"? Make up your mind, do you have k of them or an infinite number of them?2010-11-30
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    I have k types.. I can take any number from any type...2010-11-30
  • 4
    Are you sure $k^i/i!$ is the right expression? Maybe you want $\binom{k+i-1}{i-1}$?2010-11-30

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