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Lời giải:
Liên hợp ta thấy:
\(2(\sqrt{n+1}-\sqrt{n})=2.\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(1)\)
\(2(\sqrt{n}-\sqrt{n-1})=2.\frac{n-(n-1)}{\sqrt{n}+\sqrt{n-1}}=\frac{2}{\sqrt{n}+\sqrt{n-1}}>\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(2)\)
Từ \((1);(2)\Rightarrow 2(\sqrt{n+1}-\sqrt{n})< \frac{1}{\sqrt{n}}< 2(\sqrt{n}-\sqrt{n-1})\)
------------------------
Áp dụng vào bài toán:
\(S=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>1+2(\sqrt{3}-\sqrt{2})+2(\sqrt{4}-\sqrt{3})+...+2(\sqrt{101}-\sqrt{100})\)
\(\Leftrightarrow S>1+2(\sqrt{101}-\sqrt{2})>18(*)\)
Và:
\(S< 1+2(\sqrt{2}-\sqrt{1})+2(\sqrt{3}-\sqrt{2})+....+2(\sqrt{100}-\sqrt{99})\)
\(\Leftrightarrow S< 1+2(\sqrt{100}-\sqrt{1})=19(**)\)
Từ $(*); (**)$ suy ra $18< S< 19$ (đpcm)
Mình đã chứng minh \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\left(n\inℕ^∗\right)\) rồi nha!
Áp dụng vào, ta được: \(\frac{1}{2\sqrt{1}}< \sqrt{1}\)
\(\frac{1}{2\sqrt{2}}< \sqrt{2}-\sqrt{1}\)
\(\frac{1}{2\sqrt{3}}< \sqrt{3}-\sqrt{2}\)
.............................
\(\frac{1}{2\sqrt{2500}}< \sqrt{2500}-\sqrt{2499}\)
\(\Rightarrow1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2500}}\)
\(< 2\left(\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2500}-\sqrt{2499}\right)\)
\(=2.50=100\)
=> ĐPCM
P/s: sai sót xin bỏ qua cho.
\(\hept{\begin{cases}\frac{2}{2\sqrt{n}}< \frac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\frac{2}{2\sqrt{n}}>\frac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{cases}}\)
Từ đây ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2\left(\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+...+\sqrt{n}-\sqrt{n-1}\right)\)
\(=2\left(\sqrt{n}-0\right)=2\sqrt{n}\)
Tương tự ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\right)\)
\(=2\left(\sqrt{n+1}-1\right)>\sqrt{n}\)
Gọi \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}=A\)là A
Có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{n}}\)
=> \(A>n.\frac{1}{\sqrt{n}}=\sqrt{n}\)(1)
Ta có: \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=2\left(\sqrt{n}+\sqrt{n-1}\right)\)
=> \(\frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
Khi đó: \(\frac{1}{\sqrt{1}}< 2\left(\sqrt{1}-\sqrt{0}\right)\)
\(\frac{1}{\sqrt{2}}< 2\left(\sqrt{2}-\sqrt{1}\right)\)
...
\(\frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
=> \(A< 2\left(\sqrt{n}-\sqrt{n-1}+...+\sqrt{1}\right)\)
=> \(A< 2\sqrt{n}\)(2)
Từ (1) và (2) => \(\sqrt{n}< A< 2\sqrt{n}\)
* t sẽ chứng minh đề thiếu điều kiện \(n>0\)
ĐKXĐ : \(n>0\) hoặc \(n< -1\)
+) Nếu \(n>0\) ta có :
\(\frac{1}{\sqrt{n^2+1}}< \frac{1}{\sqrt{n^2}}=\frac{1}{\left|n\right|}=\frac{1}{n}\)
\(\frac{1}{\sqrt{n^2+2}}< \frac{1}{n}\)
\(\frac{1}{\sqrt{n^2+3}}< \frac{1}{n}\)
\(............\)
\(\frac{1}{\sqrt{n^2+n}}< \frac{1}{n}\)
\(\Rightarrow\)\(P=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\frac{1}{\sqrt{n^2+3}}+...+\frac{1}{\sqrt{n^2+n}}>\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}\)
\(=n.\frac{1}{n}=1\)
\(\Rightarrow\)\(P< 1\)
+) Nếu \(n< -1\) ta có :
\(\frac{1}{\sqrt{n^2+1}}< \frac{1}{\sqrt{n^2}}=\frac{1}{\left|n\right|}=\frac{1}{-n}\)
\(\frac{1}{\sqrt{n^2+2}}< \frac{1}{-n}\)
\(\frac{1}{\sqrt{n^2+3}}< \frac{1}{-n}\)
\(............\)
\(\frac{1}{\sqrt{n^2+n}}< \frac{1}{-n}\)
\(\Rightarrow\)\(P=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\frac{1}{\sqrt{n^2+3}}+...+\frac{1}{\sqrt{n^2+n}}< \frac{1}{-n}+\frac{1}{-n}+\frac{1}{-n}+...+\frac{1}{-n}\)
\(=n.\frac{1}{-n}=-1\)
\(\Rightarrow\)\(P< -1\)
Vậy nếu \(n>0\) thì \(P< 1\) , nếu \(n< -1\) thì \(P< -1\)
hehe :))
tuyệt :v