CMR : 1 + $ frac{1}{√2}$ + $ frac{1}{√3}$ + $ frac{1}{√4}$ + ... + $ frac{1}{√n}$ √n ( n 1 )

Question

CMR : 1 +  $ frac{1}{√2}$ + $ frac{1}{√3}$ + $ frac{1}{√4}$ + ... + $ frac{1}{√n}$  > √n ( n > 1 )

uyenthu · Answer

Lời giải:    Ta c&oacute; `:`   `+)` `1/sqrt1>1/sqrtn``text(     (1))`   `+)` `1/sqrt2>1/sqrtn``text(     (2))`   `............`   `+)` `1/sqrtn=1/sqrtn``text(     (n))`   Từ tr&ecirc;n suy ra `:`   `1/sqrt1+1/sqrt2+1/sqrt3+...+1/sqrtn>1/sqrtn.n=sqrtn`   Vậy `1/sqrt1+1/sqrt2+1/sqrt3+...+1/sqrtn>sqrtn`

huyenmi · Answer

ta c&oacute; : `1/(&radic;1)>1/(&radic;n)` v&igrave; `( n > 1 )`   `1/(&radic;2)>1/(&radic;n)` v&igrave; `( n > 1 )`   `1/(&radic;3)>1/(&radic;n)` v&igrave; `( n > 1 )`   `1/(&radic;4)>1/(&radic;n)` v&igrave; `( n > 1 )`   cứ tiếp tục như vậy :   `1/(&radic;n -1)>1/(&radic;n)` v&igrave; `( n > 1 )`   `1/(&radic;n)=1/(&radic;n)` v&igrave; `( n > 1 )`   `&rArr;1/(&radic;1)+1/(&radic;2)+1/(&radic;3)+1/(&radic;4)+....+1/(&radic;n -1)+|1/(&radic;n)>1/(&radic;n)+1/(&radic;n)+1/(&radic;n)+1/(&radic;n)+....+1/(&radic;n)`   `&rArr;1/(&radic;1)+1/(&radic;2)+1/(&radic;3)+1/(&radic;4)+....+1/(&radic;n -1)+|1/(&radic;n)>(n)/(&radic;n)`   `&rArr;1/(&radic;1)+1/(&radic;2)+1/(&radic;3)+1/(&radic;4)+....+1/(&radic;n -1)+|1/(&radic;n)>&radic;n`

CMR : 1 + $\frac{1}{√2}$ + $\frac{1}{√3}$ + $\frac{1}{√4}$ + … + $\frac{1}{√n}$ > √n ( n > 1 )

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