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a, Chắc xét hàm số tổng quát!
Xét hàm số tổng quát:
\(\dfrac{1}{\left(k+1\right)\sqrt{k}}=\dfrac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\dfrac{1}{k\left(k+1\right)}\right)\)
\(=\sqrt{k}\left[\sqrt{\dfrac{1}{k}}^2-\sqrt{\dfrac{1}{k+1}}^2\right]\)
\(=\sqrt{k}\left(\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(=\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
Vì \(\dfrac{\sqrt{k}}{\sqrt{k+1}}< 1\Rightarrow1+\dfrac{\sqrt{k}}{\sqrt{k+1}}< 2\)
Do đó \(\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)< 2.\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(\Rightarrow\dfrac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\) (1)
Áp dụng điểu (1) ta được:
\(\dfrac{1}{2}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\right)\)
\(\dfrac{1}{3\sqrt{2}}< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)\)
...................................
\(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Với mọi giá trị của \(n>0\) ta luôn có: \(\sqrt{n+1}>0\)
Do đó \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\) (đpcm)
Vì n là số tự nhiên nên ta có:
\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\ge1\)
CMR n\(\in \)N, n>3
\(\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{n^3} <2 \)
@Akai Haruma
Với \(n\ge3\) thì ta có:
\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-2\right)\left(n-1\right)n}=\dfrac{1}{2}\left(\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right)\)
Áp dụng vào bài toán ta được
\(\dfrac{1}{1^3}+\dfrac{1}{2^3}+...+\dfrac{1}{n^3}\)
\(< 1+\dfrac{1}{8}+\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right)\)
\(=1+\dfrac{1}{8}+\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{\left(n-1\right)n}\right)\)
\(< 1+\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{11}{8}< 2\)
\(\dfrac{1}{2!}+\dfrac{5}{3!}+\dfrac{11}{4!}+\dfrac{19}{5!}+...+\dfrac{n^2+n-1}{\left(n+1\right)!}\)
\(=\dfrac{1}{2!}+\dfrac{2^2+2-1}{\left(2+1\right)!}+\dfrac{3^2+3-1}{\left(3+1\right)!}+\dfrac{4^2+4-1}{\left(4+1\right)!}+...+\dfrac{n^2+n-1}{\left(n+1\right)!}\)
\(=\dfrac{1}{2!}+\dfrac{2.\left(2+1\right)-1}{\left(2+1\right)!}+\dfrac{3.\left(3+1\right)-1}{\left(3+1\right)!}+\dfrac{4.\left(4+1\right)-1}{\left(4+1\right)!}+...+\dfrac{n.\left(n+1\right)-1}{\left(n+1\right)!}\)
\(=\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+\dfrac{1}{3!}-\dfrac{1}{5!}+...+\dfrac{1}{\left(n-1\right)!}-\dfrac{1}{\left(n+1\right)!}\)
\(=\dfrac{1}{2!}+\left(\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{\left(n-1\right)!}\right)-\left(\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{5!}+...+\dfrac{1}{\left(n+1\right)!}\right)\)
\(=\dfrac{1}{2!}+\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{n!}-\dfrac{1}{\left(n+1\right)!}\)
\(=2-\dfrac{n+1+1}{\left(n+1\right)!}\)
\(=\dfrac{2\left(n+1\right)!-n-2}{\left(n+1\right)!}\)
@soyeon_Tiểubàng giải