Salut,
963:
1 ≤ k ≤ n => 1 ≤ k² ≤ n² |+1, deci 2 ≤ k² + 1 ≤ n² + 1.
[tex]\sqrt2\leqslant\sqrt{k^2+1}\leqslant\sqrt{n^2+1}\ |+n\Rightarrow n+\sqrt2\leqslant n+\sqrt{k^2+1}\leqslant n+\sqrt{n^2+1}\Rightarrow\\\Rightarrow\dfrac{1}{n+\sqrt{n^2+1}}\leqslant\dfrac{1}{n+\sqrt{k^2+1}}\leqslant\dfrac{1}{n+\sqrt2}\Bigg{|}\sum\Rightarrow\\\\\Rightarrow\sum\limits_{k=1}^n\dfrac{1}{n+\sqrt{n^2+1}}\leqslant\sum\limits_{k=1}^n\dfrac{1}{n+\sqrt{k^2+1}}\leqslant\sum\limits_{k=1}^n\dfrac{1}{n+\sqrt2}\Rightarrow\\\\\Rightarrow\dfrac{n}{n+\sqrt{n^2+1}}\leqslant\sum\limits_{k=1}^n\dfrac{1}{n+\sqrt{k^2+1}}\leqslant\dfrac{n}{n+\sqrt2}.[/tex]
Cele 2 limite laterale sunt 1, deci conform criterului cleștelui, limita din enunț este egală cu 1.
Simplu, nu ?
Green eyes.
