Răspuns:
[tex]n^2 < n(n+1) < (n+1)^2\Rightarrow n < \sqrt{n(n+1)} < n+1\Rightarrow [\sqrt{n(n+1)}]=n[/tex]
[tex]x_n=\sqrt{n(n+1)}-[\sqrt{n(n+1)}]=\sqrt{n(n+1)}-n[/tex]
[tex]\displaystyle\lim_{n\to\infty}\left(\sqrt{n(n+1)}-n\right)=\lim_{n\to\infty}\dfrac{n}{\sqrt{n(n+1)}+n}=\lim_{n\to\infty}\dfrac{n}{n\left(\sqrt{1+\frac{1}{n}}+1}\right)}=\dfrac{1}{2}[/tex]
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Răspuns:
[tex]n^2 < n(n+1) < (n+1)^2\Rightarrow n < \sqrt{n(n+1)} < n+1\Rightarrow [\sqrt{n(n+1)}]=n[/tex]
[tex]x_n=\sqrt{n(n+1)}-[\sqrt{n(n+1)}]=\sqrt{n(n+1)}-n[/tex]
[tex]\displaystyle\lim_{n\to\infty}\left(\sqrt{n(n+1)}-n\right)=\lim_{n\to\infty}\dfrac{n}{\sqrt{n(n+1)}+n}=\lim_{n\to\infty}\dfrac{n}{n\left(\sqrt{1+\frac{1}{n}}+1}\right)}=\dfrac{1}{2}[/tex]
Explicație pas cu pas: