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I observed for the function $$ f(n)= e^n \sum_{k=0}^{n-1}\left(\dfrac{k - n}{e}\right)^k \cdot \dfrac{1}{k!} \tag 1$$

with small $n$ that

  n  sum
 -------------
  1  2.7182818
  2  4.6707743
  3  6.6665656
  4  8.6666045
  5  10.666662
  6  12.666667
  7  14.666667
  8  16.666667

So an obvious hypothesis is $$ \lim_{n \to \infty} \bigl(f(n)-2n\bigr) = \frac 23 \tag 2$$

However, I have no idea, how to prove this but would like to understand how I can approach such a proof (I'll have then some similar ones with likely the same or related logic)

So I would like to understand ...

Q: how I could prove that assumed limit (2).

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the title is certainly wrong: you can't have a limit in $n$ which is a function of $n$. –  Alex Jun 23 at 7:27
2  
@Alex, Just consider $$\lim_{n \to \infty} -2n + e^n \sum_{k=0}^{n-1} ({k-n \over e})^k/k! = \frac 23$$ –  Fabien Jun 23 at 7:31
    
what you wrote may be correct. The title is certainly not. –  Alex Jun 23 at 7:33
    
@Alex: I got it, thanks! Just edited the title. –  Gottfried Helms Jun 23 at 7:54

3 Answers 3

This is a pretty neat problem actually so I'm only giving you a little hint. Start by rewriting as $$\sum_{k=0}^{n-1}\frac{1}{k!}(k-n)^{k}e^{-(k-n)}=\sum_{k=0}^{n-1}\frac{1}{k!}\frac{d^{k}}{dx^{k}}e^{x(k-n)}|_{x=-1}=\sum_{k=0}^{n-1}\frac{1}{2\pi i}\int_{C}\frac{e^{z(k-n)}}{(z+1)^{k+1}}dz$$

Since $e^{z(k-n)}$ is analytic in all of $\mathbb{C}$ I applied Cauchys Integral formula and $C$ denotes an appropriate cirle that encloses the multiple singularity $z=-1$

Now feel free to expand further to a geometric sum and you shouldn't be too far from a final solution.

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Very neat -thank you! Only I'm illiterate with that type of integral so I do not know how to proceed here (and more: I doubt I can use/adapt this idea by myself to the other cases in my set of limit-epressions). I know I can look in wikipedia (and have done this from time to time earlier) for the Cauchy integral or circular/path integral but I've never got the key idea of it and how to operate with it actually... –  Gottfried Helms Jun 23 at 12:18
1  
So after bringing the sum inside and simplifying we are left with $\frac{1}{2\pi i} \int_C \frac{dz}{(1+z)^n(e^z - z - 1)}$. How do you see that this is close to $2n + \frac{2}{3}$? –  J. J. Jun 23 at 12:30
    
Perhaps it is useful to know, that I got that $f(n)$ as partial sums (always up to $n$) of a transform of the geometric series (with $q=1$), which must somehow be related to the Borel-summation method. Also in the analysis of the Borel-summation I've seen the consideration of an integral, however I don't know whether it is related. But perhaps this loose relation contains another clue? –  Gottfried Helms Jun 23 at 13:42

One can notice that your formula is the expected number of $[0,1]$-uniformly distributed random variables that are needed for their sum to exceed $n$. (See http://mathworld.wolfram.com/UniformSumDistribution.html.) Moreover, the question http://mathoverflow.net/questions/141368/error-term-for-renewal-function discusses the behaviour of the error term $\epsilon(n) = f(n) - 2n - 2/3$.

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Wow, what an incidence (the MO-question)! The formula occurs by the matrix-summation-method based on the "Eulerian numbers" and of which I've asked&discussed several aspects (also here and in MO). Now it would be interesting whether also the generalizations might have similar relatives like that in the MO-question... –  Gottfried Helms Jun 23 at 15:16
    
I've just put some background-information from my work with this at the MO-site's question, which you've linked to. Perhaps this is interesting,too –  Gottfried Helms Jun 24 at 2:40

Notice that $$f'(n) = f(n) - f(n-1)$$

so this is a simple (but special) case of a delay differential equation. Simply substituting $f(n) = e^{an}$ reveals the form of the general solution: $$\sum_{-\infty}^{\infty} c_k e^{(1 + W_k(-\frac{1}{e}))n}$$

where $W_k(x)$ is the k-th branch of the LambertW function, and $c_k$ are parameters depending on the boundary conditions.

To get the solution we must take into account the double root at $k=-1,0$, where $W_{-1}(-\frac{1}{e}) = W_0(-\frac{1}{e}) = -1$:

$$f(n) = (c_{-1} +c_0n)e^0 + \sum_{-\infty,\neq 0,-1}^{\infty} c_k e^{(1 + W_k(-\frac{1}{e}))n}$$

but from definition $$e^{1+W_k(-\frac{1}{e})} = -\frac{1}{W_k(-\frac{1}{e})}$$ and $|\!|W_k(-\frac{1}{e})|\!| > 1$ for $k\neq, 0, -1$, so

$$\lim_{n\to\infty} f(n) = c_{-1} + c_0n + 0$$

See here for a solution that uses Laplace transform.

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