Chứng minh rằng: \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}<2\)
a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
Chứng minh rằng\(\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< \sqrt{2}\)
Chứng minh rằng: \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Ta có : \(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\frac{1}{k\left(k+1\right)}\right)=\sqrt{k}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\sqrt{k}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+}}\right)\)
\(=\left(1+\frac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
Áp dụng : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+}}< 2\)
Vậy ta có điều phải chứng minh.
Chứng minh rằng
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)
Chứng minh rằng với mọi số nguyên dương n ta đều có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\frac{1}{5\sqrt{4}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Xét số hạng tổng quát ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\sqrt{n}\cdot\frac{2}{\sqrt{n}}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài tập, ta có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+...+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
\(=2-\frac{2}{\sqrt{n+1}}< 2\left(đpcm\right)\)
Chứng minh rằng: \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\) với mọi số nguyên dương n
Chứng minh:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< 1-\frac{1}{\sqrt{n+1}}\left(n\in N\right)\)
Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)
\(< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}-\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta có:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
.....
\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)
Chứng minh rằng
\(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)với \(n\inℕ^∗\)
Áp dụng cho \(S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
chứng minh rằng 18<S<19