Răspuns:
[tex]\lim _{n\to \infty }\left(\frac{1}{\sqrt{n^2}+1}+\frac{1}{\sqrt{n\:^2+2}}+...+\frac{1}{\sqrt{n\:^2+n}}\right)=0[/tex]
Explicație pas cu pas:
[tex]\lim _{n\to \infty }\left(\frac{1}{\sqrt{n^2}+1}+\frac{1}{\sqrt{n\:^2+2}}+...+\frac{1}{\sqrt{n\:^2+n}}\right)\\\\\lim _{n\to \infty }\left(\frac{1}{\sqrt{\infty ^2}+1}+\frac{1}{\sqrt{\infty \:^2+2}}+...+\frac{1}{\sqrt{\infty \:^2+\infty }}\right)[/tex]
atunci cand numitorul unei fractii contine simbolul ∞, atunci fractia este egala cu 0 .
[tex]\lim _{n\to \infty }(0+0+...+0)\\\lim _{n\to \infty }\left(\frac{1}{\sqrt{n^2}+1}+\frac{1}{\sqrt{n\:^2+2}}+...+\frac{1}{\sqrt{n\:^2+n}}\right)=0[/tex]
Succes! :)
