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Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
....................
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(=2-\frac{1}{n}\)
đpcm
Tham khảo nhé~
Ta có :
\(\hept{\begin{cases}\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\end{cases}}\forall n\in N\)
Suy ra : \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)
Đặt \(M=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2499}}+\frac{1}{\sqrt{2500}}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{2\sqrt{2500}}+\frac{1}{2\sqrt{2499}}+...+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{2}}+\frac{1}{2}\)
Áp dụng BĐT , ta có :
\(\frac{1}{2}M< \sqrt{2500}-\sqrt{2499}+\sqrt{2499}-\sqrt{2498}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}M< \sqrt{2500}-\sqrt{1}+\frac{1}{2}=50-\frac{1}{2}< 50\)
\(\Rightarrow M< 100\)
Tham khảo nè:
1/2^2 + 1/3^2 + 1/4^2 + ... + 1/n^2 < 2/3 chứng minh
k² > k² - 1 = (k-1)(k+1)
⇒ 1/k² < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*)
Áp dụng (*), ta có:
1/2² + 1/3² + 1/4² + ... + 1/n²
< 1/2² + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)]
= 1/2² + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2
= 1/2² + [1/2 + 1/3 - 1/n - 1/(n+1)]/2
= 2/3 - [1/n + 1/(n+1)]/2 < 2/3