Question

Cho dãy số (\(U_n\)) với u_n=\(\dfrac{1}{\sqrt{n^2+1}}\)+\(\dfrac{1}{\sqrt{n^2+2}}\)+...+\(\dfrac{1}{\sqrt{n^2+n}}\)                   Tính \(\overset{lim}{n\rightarrow+\infty}\)u_n

Answer 1

Có: \(u_n=\sum\limits^n_{k=1}\dfrac{1}{\sqrt{n^2+k}}< \dfrac{n}{\sqrt{n^2+n}}=\dfrac{1}{\sqrt{1+\dfrac{1}{n}}}\)

Lại có: \(u_n=\sum\limits^n_{k=1}\dfrac{1}{\sqrt{n^2+k}}>\dfrac{n}{\sqrt{n^2+1}}=\dfrac{1}{\sqrt{1+\dfrac{1}{n^2}}}\)

\(\Rightarrow\dfrac{1}{\sqrt{1+\dfrac{1}{n^2}}}< u_n< \dfrac{1}{\sqrt{1+\dfrac{1}{n}}},\forall n\ge1\)

Mặt khác \(\lim\limits\left(\dfrac{1}{\sqrt{1+\dfrac{1}{n^2}}}\right)=\lim\limits\left(\dfrac{1}{\sqrt{1+\dfrac{1}{n}}}\right)=1\), nên theo nguyên lí kẹp, ta có \(limu_n=1\)