Question

Calculati limitele celor 2 siruri, cand n tinde la infinit:

[tex]\{\sqrt{n^2+n}\}\ \ , n\geq1\\
\{\sqrt{n^2+5n}\}\ \ , n\geq1[/tex]

Answer 1

[tex]\it n^2 \ \textless \ n(n+1) \ \textless \ (n+1)^2 \Rightarrow \sqrt{n^2} \ \textless \ \sqrt{n(n+1)} \ \textless \ \sqrt{(n+1)^2} \Rightarrow \\\;\\ \Rightarrow n \ \textless \ \sqrt{n(n+1)} \ \textless \ n+1 \Rightarrow [\sqrt{n(n+1)}] = n[/tex]


[tex]\it \{\sqrt{n(n+1)}\} = \sqrt{n(n+1)} -n = \dfrac{(\sqrt{n(n+1)} -n)(\sqrt{n(n+1)} +n)}{\sqrt{n(n+1)} +n} \\\;\\ \\\;\\ = \dfrac{n^2+n-n^2}{\sqrt{n^2+n} +n} =\dfrac{n}{n\left(\sqrt{1+\dfrac{1}{n}}+1 \right)} =\dfrac{1}{\sqrt{1+\dfrac{1}{n}}+1} \longrightarrow \dfrac{1}{2} =0,5[/tex]

Answer 2

..................................................