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I've found a lot of series expansions of the Napier's constant. I was wondering if a general form for this could be devised. They all have a trend and similarities. I've been trying but I've been screwing up as usual.

  1. $\Large{\sum\limits_{n=0}^{\infty} \frac{1}{(n - k)!} = e}$

  2. $\Large{\sum\limits_{n=1}^{\infty} \frac{1}{(n - k)!} = e - 1}$

where $k$ belongs to the set of natural numbers.

  1. $\Large{\sum\limits_{n = 2}^{\infty} \frac{1}{n!} = e - 2}$

  2. $\Large{\sum\limits_{n = 0}^{\infty} \frac{1}{(n + 1)!} = e - 1}$

  3. $\Large{\sum\limits_{n = 1}^{\infty} \frac{1}{(n + 1)!} = \sum\limits_{n = 0}^{\infty} \frac{1}{(n + 2)!} = e - 2}$

  4. $\Large{\sum\limits_{n = 0}^{\infty} \frac{n}{n!} = e}$

  5. $\Large{\sum\limits_{n = 1}^{\infty} \frac{n^2}{n!} = \sum\limits_{n = 1}^{\infty} \frac{n^2}{n!} = 2e }$

  6. $\Large{\sum\limits_{n = 1}^{\infty} \frac{n^3}{n!} = \sum\limits_{n = 1}^{\infty} \frac{n^3}{n!} = 5e }$

  7. $\Large{\sum\limits_{n = 1}^{\infty} \frac{n^4}{n!} = \sum\limits_{n = 1}^{\infty} \frac{n^4}{n!} = 15e }$

I was wondering if you guys could help me out and give me a general equation that encapsulates all of this. A brief description of how it works might also be of interest to me.

Thank you, in advance, for your efforts.

user642796
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Nick
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    The first two work, I presume, under the assumptions that $;k\in\Bbb N;$ and $;m!=0;$ if $;0>m\in\Bbb Z;$ ...and the most general form is, imo, $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ – DonAntonio Dec 26 '13 at 12:17
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    Does this mean that $$\sum_{n=1}^{\infty} \frac{n^k}{n!} = e \cdot B(k)$$ where B(k) is the number of set partitions of $k$? – Mitch Dec 26 '13 at 13:53
  • @Mitch: Yes ! – Lucian Dec 26 '13 at 16:38

3 Answers3

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$$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$

example 7): $$\sum_{n=1}^{\infty}\frac{n^{2}}{n!}=\sum_{n=0}^{\infty}\frac{n+1}{n!}=\sum_{n=0}^{\infty}\frac{n}{n!}+e=2e$$ where 6) is used

drhab
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You didn't specify what $k$ is in the first two descriptions but the correct forms would be the ones with $k=0$. All your formulas can be derived from the first: $$ e=\sum_{n=0}^\infty \frac{1}{n!}=1+1+\frac{1}{2}+\ldots. $$ Number 2 omits the first summand in the series so the result is $e-1$. Number 3 omits the first and the second. Number 4 is just an index shift away from number 2. I'll do one of the others: $$ \sum_{n=1}^\infty \frac{n^2}{n!}=\sum_{n=1}^\infty \frac{n}{(n-1)!}=\sum_{n=0}^\infty\frac{n+1}{n!}=\sum_{n=0}^\infty\frac{n}{n!}+\sum_{n=0}^\infty\frac{1}{n!}=e+e. $$

MichalisN
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the formula $4,5,6,7$ can be proved the following way. $$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$$ differentiate with respect to $x$ $$\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}=e^x$$ multiply both side by $x$ $$\sum_{n=0}^{\infty}\frac{nx^{n}}{n!}=xe^x$$ repeat this cycle of differentiating with respect to $x$ and multiplying with $x$ $3$ more times to get. $$\sum_{n=0}^{\infty}\frac{n^4x^{n}}{n!}=xe^x(x^3+6x^2+7x+1)$$ now substitute $x=1$ to get $$\sum_{n=0}^{\infty}\frac{n^4}{n!}=15e$$

Suraj M S
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