bạn ơi bài này có trong bùi văn tuyên
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}< 1\)
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}\)
\(A< \frac{99}{100}< 1\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}\text{ ko phải là 1 số tự nhiên ( đpcm )}\)
Ta có: \(\frac{1}{2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3}\)< \(\frac{1}{2.3}\)
...........................
\(\frac{1}{100}\)< \(\frac{1}{99.100}\)
Suy ra A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ .....+ \(\frac{1}{100}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+......+\(\frac{1}{99.100}\)
Suy ra A=\(\frac{1}{2}\)+\(\frac{1}{3}\)+...........+\(\frac{1}{100}\)<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.........+\(\frac{1}{99}\)-\(\frac{1}{100}\)
Suy ra A= \(\frac{1}{2}\)+\(\frac{1}{3}\)+............+\(\frac{1}{100}\)< 1-\(\frac{1}{100}\)= \(\frac{99}{100}\)
Vì \(\frac{99}{100}\)<1 mà A= \(\frac{1}{2}\)+\(\frac{1}{3}\)+......+\(\frac{1}{100}\)< \(\frac{99}{100}\)
Nên A không là số tự nhiên(đpcm)