Limite d'une suite

supp

supp

Bonjour à tous
je vous propose ce raisonnement pour calculer la limite de la suite
$ \Large u_{n}= \frac{n}{n^{3}+1}+ \frac{2n}{n^{3}+2}+...+\frac{n.n}{n^{3}+n}$
on a :
$

\Large


u_{n} = \frac{1}{n^{2}}.
\sum_{k=1}^{n} \frac{k}{1+\frac{k}{n^{3}}} \\
sachant~~ que:
\frac{1}{1+\frac{k}{n^{3}}} = \int_{\frac{1}{2}}^{1+\frac{k}{n^{3}}} \frac{-dx}{x^{2}} ~~+2\\
alors: ~~ u_{n}= \frac{1}{n^{2}}.\left(
\sum_{k=1}^{n} k\left(\int_{\frac{1}{2}}^{1} \frac{-dx}{x^{2}}+2\right)
+\varepsilon_{n} \right) \\
\\
avec:\varepsilon_{n}=\sum_{k=1}^{n} k.\int_{1}^{1+\frac{k}{n^{3}}} \frac{-dx}{x^{2}}\\
or: \int_{1}^{1+\frac{k}{n^{3}}} \frac{-dx}{x^{2}}= \frac{1}{1+\frac{k}{n^{3}}}-1\\
puisque: 1\leq k \leq n ~~et ~~ \sum_{k=1}^{n}k = \frac{n(n+1)}{2}\\
Alors: \frac{-k}{n^{3}+1} \leq k.\int_{1}^{1+\frac{k}{n^{3}}} \frac{-dx}{x^{2}} \leq \frac{-k}{n^{2}+1}\\
\\
et :~~ \frac{-n(n+1)}{2(n^{3}+1)} \leq \varepsilon_{n}
\leq \frac{-n(n+1)}{2(n^{2}+1)}\\
d'ou:~~~~\lim\limits_{n \rightarrow +\infty} \frac{\varepsilon_{n}}{n^{2}}=0\\
or: u_{n}=\frac{1}{n^{2}}\left(
\int_{\frac{1}{2}}^{1} \frac{-dx}{x^{2}}+2
\right)
\sum_{k=1}^{n}k+ \frac{\varepsilon_{n}}{n^{2}}\\
c.à.d : ~~ u_{n}=\frac{n(n+1)}{2n^{2}}\left(
\int_{\frac{1}{2}}^{1} \frac{-dx}{x^{2}}+2
\right)
+ \frac{\varepsilon_{n}}{n^{2}}\\
or : ~~ \int_{\frac{1}{2}}^{1} \frac{-dx}{x^{2}}=-1\\
Alors:
\lim\limits_{n \rightarrow +\infty} u_{n}=\frac{1}{2}\\
$