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I wish to find sum of a finite and infinite series$$\sum \frac{1}{n!}$$

I am aware, that this is a standard series and thus has a straight forward (well-known) answer BUT I am not recollecting it.

A hint Please.

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3  
$e$ –  anon Jun 28 '12 at 9:32
1  
for infinite series, it is $e$, otherwise i am afraid there is no closed form available. –  Aang Jun 28 '12 at 9:34

2 Answers 2

up vote 2 down vote accepted

Do you remember that $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\ ?$$

Edit: This is the answer for the infinite series.

Edit: The convergence is easy: just apply the ratio test.

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Actually I just want to show the series is convergent. I would not like to go for another more complicated series. Better if a simple and short answer is available –  rajendra bakre Jun 28 '12 at 9:43
    
Try using the ratio test. –  Benji Jun 28 '12 at 10:23

Since you have asked for finite sums too:

$$ \sum_{n=0}^m 1/n! = \frac{e\;\Gamma(m+1,1)}{\Gamma(m+1)}, $$ where $\Gamma(m+1,x) = \int_x^{\infty} t^{m}\,e^{-t}\,{\rm d}t \,$ resp. $\Gamma(m+1)= \int_0^{\infty} t^{m}\,e^{-t}\,{\rm d}t $ are the (incomplete) $\Gamma$ functions.

Further you'll find in limit, that

$$ \begin{eqnarray} \Gamma(s,x) &=& (s-1)!\, e^{-x} \sum_{k=0}^{s-1} \frac{x^k}{k!}\\ \frac{\Gamma(s,x)}{\Gamma(s)}&=&e^{-x} \sum_{k=0}^{s-1} \frac{x^k}{k!}\\ \lim_{s\to \infty} e^{-x} \sum_{k=0}^{s-1} \frac{x^k}{k!}&=&e^{-x}e^x=1, \end{eqnarray} $$ stated by the other answers given.

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