Question

[Khánh Phạm]

Cho F = 1/2^2 + 1/3^2 + 1/4^2 + .... + 1/100^2. Chứng minh 7/15 < F < 4/3

Answer 1

[Nguyễn Thị Thương Hoài]

F = \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + ... + \(\frac{1}{100^2}\)

\(\frac12-\frac13\) = \(\frac{1}{2.3}\) < \(\frac{1}{2^2}\) < \(\frac{1}{1.2}\) = \(\frac11\) - \(\frac12\)

\(\) \(\frac13-\frac14\) = \(\frac{1}{3.4}\) < \(\frac{1}{3^2}\) < \(\frac{1}{2.3}\) = \(\frac12-\frac13\)

..............................................................

\(\frac{1}{100}-\frac{1}{101}\) = \(\frac{1}{100.101}\) < \(\frac{1}{100^2}\) < \(\frac{1}{99.100}\) = \(\frac{1}{99}-\frac{1}{100}\)

Cộng vế với vế ta có:

\(\frac12-\frac{1}{101}\) < \(\frac{1}{2^2}+\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ ... + \(\frac{1}{100^2}\) < \(\frac11\) - \(\frac{1}{100}\) < 1 < \(\frac43\)

\(\frac12\) - \(\frac{1}{101}\) < F < \(\frac34\)

\(\frac12-\frac{1}{30}\) < \(\frac12\) - \(\frac{1}{101}\) < F < \(\frac34\)

\(\frac{7}{15}\) < F < \(\frac34\)

Answer 2

[Nguyễn Hoàng Anh Thư]

Bất đẳng thức 1: F > 7/15 Ta có:
F = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2

1/2^2 + 1/34 + 1/45 + ... + 1/100*101
= 1/4 + (1/3 - 1/4) + (1/4 - 1/5) + ... + (1/100 - 1/101)
= 1/4 + 1/3 - 1/101
1/4 + 1/3 - 1/100
= 7/12 - 1/100
7/15

Bất đẳng thức 2: F < 4/3 Ta có:
F = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
< 1/12 + 1/23 + 1/34 + ... + 1/99100
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100)
= 1 - 1/100
< 1 + 1/3
= 4/3 Vậy 7/15 < F < 4/3.