Question

Chứng minh rằng:

a=1+1/2^2+1/3^2+1/4^2+...+1/100^2<2

b=1+1/2+1/3+1/4+...+1/63<6

c=1/2.3/4.5/6....9999/10000<1/100

Answer 1

a, Ta có : \(\dfrac{1}{2^2}=\dfrac{1}{4};\dfrac{1}{3^2}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{4^2}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(...\dfrac{1}{100^2}< \dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}< 2\)

@Nguyễn Khanh

Answer 2

b, 1 = 1
1/2 + 1/3 = 1/(1 + 1) + 1/(1 + 2) < 2/(1 + 1) = 2/2 = 1
1/4 + 1/5 + 1/6 + 1/7 = 1/(3 + 1) + 1/(3 + 2) + 1/(3 + 3) + 1/(3 + 4) < 4/(3 + 1) = 4/4 = 1
1/8 + 1/9 + ... + 1/15 = 1/(7 + 1) + 1/(7 + 2) + ... + 1/(7 + 8) < 8/(7 + 1) = 8/8 = 1
1/16 + 1/17 + ... + 1/31 = 1/(15 + 1) + 1/(15 + 2) + ... + 1/(15 + 16) < 16/(15 + 1) = 16/16 = 1
1/32 + 1/33 + ... + 1/63 = 1/(31 + 1) + 1/(31 + 2) + ... + 1/(31 + 32) < 32/(31 + 1) = 32/32 = 1
=> 1 + 1/2 + 1/3 + 1/4 + ... + 1/64 < 1 + 1 + 1 + 1 + 1 + 1
=> 1 + 1/2 + 1/3 + 1/4 + ... + 1/64 < 6 (đpcm)
@Nguyễn Khanh

Answer 3

Đặt A = 1/2 . 3/4 . 5/6 . ... . 9999/10000 (A > 0)
Và B = 2/3 . 4/5 . 6/7 . ... . 10000/10001 (B > 0)
Ta có A.B = 1/2 . 2/3 . 3/4 . ... . 10000/10001 = 1/10001 (1)
Mặt khác :
1/2 < 2/3
3/4 < 4/5
...
9999/10000 < 10000/10001
Nhân tất cả theo vế ---> A < B ---> A² < A.B (2)
Từ (1),(2) => A² < 1/10001 => A < \(\sqrt{\dfrac{1}{10001}}\) < \(\sqrt{\dfrac{1}{10000}}\) = 1/100 (đpcm)

@Nguyễn Khanh