* 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 :))
