Timeline for Approximating finite sum of factorial reciprocal
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 7, 2021 at 16:01 | comment | added | PierreCarre | @a6623 Please consider marking as correct the answer by Claude Leibovici. Note that the formula he presented, although derived using the $\Gamma$ function, does not actually use it in the calculations. For instance, $$ \sum_{k=1}^{10}\frac{1}{k!} = \frac{\lfloor e 10!\rfloor}{10!} -1 = \frac{6235301}{3628800} $$ This is an exact result. It can't get any better! | |
| Dec 7, 2021 at 15:53 | comment | added | PierreCarre | If $f\in c^{N+1}$ in some open set centred in $a$ and containing $x$, we have that $$ f(x) = f(a) + f'(a)(x-a) + \cdots + \frac{f^{(N)}(a)}{N!}(x-a)^N + \frac{f^{(N+1)}(\xi)}{(N+1)!}(x-a)^{N+1}, \quad \xi \in (a,x). $$ In this case, since $a=0$, $x=1$ and $f^{(N+1)}(x) = e^x$, we have that $$ \left| \dfrac{f^{(N+1)}(\xi)}{(N+1)!}(1-0)^{N+1}\right| = \dfrac{e^{\xi}}{(N+1)!}\leq \frac{e}{(N+1)!} $$ | |
| Dec 7, 2021 at 15:13 | comment | added | user983799 | Thanks for your answer, is it possible you could elaborate on how $\frac{e^{\xi}}{(N+1)!}$ is derived from the taylor formula? | |
| Dec 7, 2021 at 14:59 | vote | accept | CommunityBot | ||
| Dec 7, 2021 at 17:01 | |||||
| Dec 7, 2021 at 9:50 | history | answered | PierreCarre | CC BY-SA 4.0 |