Chứng tỏ $\frac{49}{100}$ $<$ $\frac{1}{2^{2}}$ $+$ $\frac{1}{3^{2}}$ $+$ $\frac{1}{4^{2}}$ $+$ $...$ $+$ $\frac{1}{100^{2}}$ $<1$ HELP ME!!!

Đáp án:

 `1/2^2+1/3^2+1/4^2+...+1/100^2`

`1/2.3<1/2^2<1/1.2`

`1/3.4<1/3^2<1/2.3`

`1/4.5<1/4^2<1/3.4`

`...........`

`1/100.101<1/100^2<1/99.100`

`=>1/2.3+1/3.4+...+1/100.101 <1/2^2+1/3^2+1/4^2+...+1/100^2`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2>1/2-1/3+1/3-1/4+....+1/100-1/101`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2>1/2-1/101=99/202>49/100`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2>49/100`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2<1/1.2+1/2.3+..+1/99.100`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2<1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2<1-1/100<1`

`<=> 1/2^2+1/3^2+1/4^2+...+1/100^2<1`

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