Đặt \(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2009^3}\)

Ta CM công thức sau :

\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)

Thật vậy ta có : \(\left(n-1\right).n.\left(n+1\right)=\left(n-1\right)\left(n+1\right).n=\left(n^2-1\right).n=n^3-n< n^3\\ \Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)

Áp dụng công thức trên vào biểu thức A ; ta có :

\(A=\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2009^3}\\ < \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2009.2010}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.2009.2010}< \dfrac{1}{4}\)

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giúp mình nhanh lên các bạn ơi

-Với n=1, ta thấy bthức đúng.

-Với n=k, có: \(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2k-1}{4+\left(2k-1\right)^4}=\frac{k^2}{4k^2+1}=\frac{1}{4}-\frac{1}{4}.\frac{1}{4k^2+1}\)

-Giả sử bthức đúng với n=k+1, có:

\(\left(\frac{1}{4}-\frac{1}{4}.\frac{1}{4\left(k+1\right)^2+1}\right)-\left(\frac{1}{4}-\frac{1}{4}.\frac{1}{4k^2+1}\right)\)

\(=\frac{1}{4}\left(\frac{1}{4k^2+1}-\frac{1}{4\left(k+1\right)^2+1}\right)\)

\(=\frac{2k+1}{\left(4k^2+1\right)\left(4\left(k+1\right)^2+1\right)}=\frac{2k+1}{4+\left(2k+1\right)^4}\)

Vậy ta có đpcm.

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{1990^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)

\(=1-\frac{1}{1990}=\frac{1989}{1990}\)

Vì \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{1990^2}< \frac{1989}{1990}< \frac{3}{4}\)nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}< \frac{3}{4}\)