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.
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.
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.
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.