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These polynomials in $e$ converge to 2$$f(i)=e^i - i \sum_{k=1}^{i-1}\frac{(i-k)^{k-1}{e^{i-k}}{(-1)^{k+1}}}{k!}, \text{ where } i>1$$

This function goes to 2. I've calculated this with sage math tool.

$$f(\infty) = 2$$ for example, $$f(2)=e^2-2e=1.95249244... $$ $$f(3)=e^3-3e^2+\frac{3}2e=1.99579136... $$ $$f(4)=e^4-4e^3+4e^2-\frac{2}3e=2.000038... $$ $$...$$ $$f(10)=\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{36288} \, e + \frac{2}{63} \, e^{2} - \frac{81}{56} \, e^{3} + \frac{128}{9} \, e^{4} - \frac{625}{12} \, e^{5} + 90 \, e^{6} - \frac{245}{3} \, e^{7} + 40 \, e^{8} - 10 \, e^{9} + e^{10} = 2.00000000...$$
Isn't this interesting?
These polynomials in e (2.71828182845904523536...) converge to 2.

However, I have no idea how to mathematically prove this.
I guess this would have been already proved, but I have no idea where I can find the proof.

I would greatly appreciate it if you can give me some tips or the proof of this convergence.

For more information, this function $f(i)$ is from a different function $h_i(x),\text{ when } x = 1$ of an original problem
$$ f(i) = h_i(1)$$ so proving the convergence of the above polynominals will be the same as proving $h_\infty(1) = 2$

I have recently found that the general form of $h_i(x)$ function $$h_i(x) = (-1)^{i+1} e^x \left[\frac{1}{(i-1)!}{x}^{i-1} - \sum_{k=1}^{i-1}\left(α(k)\frac{{x}^{i-1-k}(-1)^{k+1}}{(i-1-k)!}\right)\right] - α(i-1),$$ where $$α(j) = \sum_{k=0}^{j-1}\frac{(j-k)^k}{k!}e^{j-k}(-1)^k$$

update list

  • I have found this: $ f(i)=α(i)−α(i−1) $
  • I have found a new property of $α(i)$, when $i > 1$, $$ α(i) = \sum_{k=0}^{i-2} \left( e \space α(k+1)\space\frac{(-1)^{i+k+2}}{(i-k-2)!} \right) + \frac{(-1)^{i+1} \space e}{(i-1)!},$$ where $α(1) = e$.
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@countinghaus thank you very much for your help. I have been struggling with this problem for a very long time. I will look forward to seeing the result. – Nate Jul 14 '12 at 14:54
If we flip the summation order, we get an equivalent expression: $$e^i+i\sum_{k=1}^{i-1} \frac{(-1)^{i-k} e^k k^{i-k-1}}{(i-k)!}$$ – J. M. Jul 14 '12 at 15:13
Your sum looks awfully like a binomial transform; I suspect this might be one way to prove the convergence of your sequence. – J. M. Jul 14 '12 at 15:41
Here is another equivalent expression: $$\frac{(-1)^i}{(i-1)!}\sum_{k=1}^i \binom{i}{k}k^{i-k}(-1)^k(k-1)!e^k$$ – J. M. Jul 14 '12 at 15:50
Terminological remark: you don't have a power series here, since the coefficients of $e^k$ depend on $i$. Every power series can be thought of as a sequence of polynomials (partial sums), but not every sequence of polynomials makes a power series. – user31373 Jul 15 '12 at 3:51
up vote 1 down vote accepted

The convergence to $2$ is explained by my answer to the original question.

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