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Soit $(a_n)$ le développement de Taylor de la fonction exponentielle comme suit:

\begin{eqnarray} a_n(x) = \sum^{n}_{k = 0} \frac{x^k}{k!} \end{eqnarray}

Soit $(y_n)$ une suite positive et strictement monotone de limite $+\infty$.

Question Peut-on prouver la limite suivante ?

\begin{eqnarray} \lim_{n \to +\infty } \frac{ a_n(y_n) }{e^{y_n}} = 1 \end{eqnarray}

Si ce n'est pas possible, peut-on prouver qu'au moins cette limite $\lim_{n \to +\infty } \frac{ a_n(y_n) }{e^{y_n}} > 0$ ?

Vous pouvez aussi trouver une version en anglais de cette question ici:Calculate the limit of a sequence: $\frac{ a_n(y_n) }{e^{y_n}}$

InfiniteMath
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    Hello Jonathan, I also created an English version of the question. Thank you very much for asking. https://math.stackexchange.com/questions/4642534/calculate-the-limit-of-a-sequence-frac-a-ny-n-ey-n – InfiniteMath Feb 20 '23 at 02:31
  • @JonathanZsupportsMonicaC: See https://math.meta.stackexchange.com/questions/19292. – joriki Feb 20 '23 at 03:39
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    @InfiniteMath: That's a bad idea, especially as you originally didn't link them together (and still haven't linked from the English version to the French). That way you're wasting everyone's time by making people who are thinking about the one version unaware of the progress made by others at the other version. Please delete this question; you're welcome to add the French version to the other question. See also https://math.meta.stackexchange.com/questions/19292 in general about posting questions in languages other than English. – joriki Feb 20 '23 at 03:42
  • @joriki - Huhn, I didn't know that. Merci pour les infos. Je vais supprimer l'au-dessus. – JonathanZ Feb 20 '23 at 03:45
  • It would be much better to write the summation as $$1+\sum_{k=1}^{n}\frac{x^k}{k!}$$. Then you can substitute $x=0$ without having to consider the illegitimate expression of 0 raised to the power 0. – P. Lawrence Feb 20 '23 at 06:48

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