CMR với mọi số nguyên dương n ta có : 1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ + … + $\frac{1}{n²}$ < $\frac{5}{3}$ 14/07/2021 Bởi Maria CMR với mọi số nguyên dương n ta có : 1 + $\frac{1}{2²}$ + $\frac{1}{3²}$ + … + $\frac{1}{n²}$ < $\frac{5}{3}$
Ta có $ a^2 > a^2 – 1 = ( a – 1).(a + 1) $ => $\frac{1}{a^2}$ < $\frac{1}{( a – 1).(a + 1)}$ = ($\frac{1}{a – 1}$ – $\frac{1}{a + 1}$ ). $\frac{1}{2}$ Áp dụng biếu thức trên ta có : $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + … + $\frac{1}{n^2}$ < $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{2.4}$ + … + $\frac{1}{( n-1)(n+1)}$ < $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + ( $\frac{1}{2}$ – $\frac{1}{4}$ + $\frac{1}{3}$ -$\frac{1}{5}$ + … + $\frac{1}{n-1}$ – $\frac{1}{n +1}$ ).$\frac{1}{2}$ < $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + ( $\frac{1}{2}$ + $\frac{1}{3}$ – $\frac{1}{n-1}$ – $\frac{1}{n }$ ).$\frac{1}{2}$ < $\frac{5}{3}$ – ($\frac{1}{n}$ + $\frac{1}{n+1}$ ) <$\frac{5}{3}$ Bình luận
Lời giải: Ta có `:` `k^2>k^2-1=(k+1)(k-1)` `⇒` `1/k^2<1/{(k-1)(k+1)}=1/2.(1/(k-1)-1/(k+1))“text( (#))` Áp dụng `(#)` ta có `:` `1+1/2^2+1/3^2+…+1/n^2<1+1/2^2+1/2.4+…+1/{(n-1)(n+1)}` `=` `1+2^2+1/2.(1/2-1/4+1/3-1/5+…+1/(n-1)-1/(n+1))` `=` `1+1/2^2+1/2.(1/2+1/3-1/n-1/(n+1))` `=` `5/3-(1/n+1/(n+1))<5/3` `->1+1/2^2+1/3^2+…+1/n^2<1+1/2^2+1/2.4+…+1/{(n-1)(n+1)}<5/3` `->` `đpcm` Bình luận
Ta có $ a^2 > a^2 – 1 = ( a – 1).(a + 1) $
=> $\frac{1}{a^2}$ < $\frac{1}{( a – 1).(a + 1)}$ = ($\frac{1}{a – 1}$ – $\frac{1}{a + 1}$ ). $\frac{1}{2}$
Áp dụng biếu thức trên ta có :
$\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + … + $\frac{1}{n^2}$ < $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{2.4}$ + … + $\frac{1}{( n-1)(n+1)}$
< $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + ( $\frac{1}{2}$ – $\frac{1}{4}$ + $\frac{1}{3}$ -$\frac{1}{5}$ + … + $\frac{1}{n-1}$ – $\frac{1}{n +1}$ ).$\frac{1}{2}$
< $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + ( $\frac{1}{2}$ + $\frac{1}{3}$ – $\frac{1}{n-1}$ – $\frac{1}{n }$ ).$\frac{1}{2}$
< $\frac{5}{3}$ – ($\frac{1}{n}$ + $\frac{1}{n+1}$ ) <$\frac{5}{3}$
Lời giải:
Ta có `:` `k^2>k^2-1=(k+1)(k-1)`
`⇒` `1/k^2<1/{(k-1)(k+1)}=1/2.(1/(k-1)-1/(k+1))“text( (#))`
Áp dụng `(#)` ta có `:`
`1+1/2^2+1/3^2+…+1/n^2<1+1/2^2+1/2.4+…+1/{(n-1)(n+1)}`
`=` `1+2^2+1/2.(1/2-1/4+1/3-1/5+…+1/(n-1)-1/(n+1))`
`=` `1+1/2^2+1/2.(1/2+1/3-1/n-1/(n+1))`
`=` `5/3-(1/n+1/(n+1))<5/3`
`->1+1/2^2+1/3^2+…+1/n^2<1+1/2^2+1/2.4+…+1/{(n-1)(n+1)}<5/3`
`->` `đpcm`