Unverified Commit a097ddb0 authored by orestis.malaspin's avatar orestis.malaspin
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corrected typo i->n

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......@@ -69,7 +69,7 @@ suivantes peuvent vous êtres utiles:
\begin{align}
\sum_{n=1}^\infty\frac{1}{n^4}=1+\frac{1}{16}+\frac{1}{81}+...=\frac{\pi}{4},\\
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=1-\frac{1}{4}+\frac{1}{9}+...=\frac{\pi^2}{12},\\
\prod_{i=1}^\infty\left(\frac{2n}{2n-1}\right)\left(\frac{2n}{2n+1}\right)=1\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot ...=\frac{\pi}{4}.
\prod_{n=1}^\infty\left(\frac{2n}{2n-1}\right)\left(\frac{2n}{2n+1}\right)=1\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot ...=\frac{\pi}{4}.
\end{align}
Quelle est la façon qui *converge* le plus rapidement?[^1] Que se passe-t-il lorsque $n$ devient trop grand?
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