Chứng minh 1/6^5 < 1/5^3 + 1/6^3

15000 câu hỏi ôn tập Toán (Phần 105)

Đề bài. Chứng minh $\frac{1}{65}<\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}<\frac{1}{40}$

Lời giải:

Ta xét bài toán tổng quát: Với mọi n thuộc ℕ^* ta có:

$\frac{1}{n^3}<\frac{1}{n^3-n}=\frac{1}{n(n^2-1)}=\frac{1}{(n-1).n.(n+1)}$

$\frac{1}{n^3}>\frac{1}{n^3+3n^2+2n}=\frac{1}{n.(n+1).(n+2)}$

Áp dụng vào bài toán ta có:

$\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}<\frac{1}{\mathrm{4.5.6}}+\frac{1}{\mathrm{5.6.7}}+\mathrm{\dots }+\frac{1}{\mathrm{2003.2004.2005}}$

$\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}<\frac{1}{2}(\frac{2}{\mathrm{4.5.6}}+\frac{2}{\mathrm{5.6.7}}+\mathrm{\dots }+\frac{2}{\mathrm{2003.2004.2005}})$

$\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}<\frac{1}{2}(\frac{1}{4.5}-\frac{1}{5.6}+\frac{1}{5.6}-\frac{1}{6.7}+\mathrm{\dots }+\frac{1}{2003.2004}-\frac{1}{2004.2005})$

$=\frac{1}{2}(\frac{1}{4.5}-\frac{1}{2003.2004})=\frac{1}{40}-\frac{1}{\mathrm{2.2003.2004}}<\frac{1}{40}$

Vậy $\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}<\frac{1}{40}$

Lại có: $\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}>\frac{1}{\mathrm{5.6.7}}+\frac{1}{\mathrm{6.7.8}}+\mathrm{\dots }+\frac{1}{\mathrm{2004.2005.2006}}$

$\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}>\frac{1}{2}(\frac{1}{5.6}-\frac{1}{6.7}+\frac{1}{6.7}-\frac{1}{7.8}+\mathrm{\dots }+\frac{1}{2004.2005}-\frac{1}{2005.2006})$

$\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}>\frac{1}{2}(\frac{1}{5.6}-\frac{1}{2005.2006})$

$\frac{1}{5^3}+\frac{1}{6^3}+\mathrm{\dots }+\frac{1}{{2004}^3}>\frac{1}{65}$