Question

Tính giới hạn của dãy số\(Un=\frac{1}{2\sqrt{1}+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

Answer 1

Lời giải:

Xét hạng tử tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{(n+1)-n}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Cho $n=1,2,...$ thì:

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=1-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

......

\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

\(\Rightarrow U_n=1-\frac{1}{\sqrt{n+1}}\)

\(\Rightarrow \lim\limits U_n=\lim (1-\frac{1}{\sqrt{n+1}})=1\)