Đặt \(A=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\)
Ta có: \(\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{n^2+n}}>...>\frac{1}{\sqrt{n^2+n}}\)
\(\Rightarrow A< \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+1}}+...+\frac{1}{\sqrt{n^2+1}}=\frac{n}{\sqrt{n^2+1}}\)
Và \(A>\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n}}=\frac{n}{\sqrt{n^2+n}}\)
\(\Rightarrow\frac{n}{\sqrt{n^2+1}}< A< \frac{n}{\sqrt{n^2+n}}\)
Mà \(lim\left(\frac{n}{\sqrt{n^2+1}}\right)=lim\left(\frac{n}{\sqrt{n^2+n}}\right)=1\)
\(\Rightarrow lim\left(A\right)=1\)
