Ta có: \(\dfrac{1}{5}=\dfrac{1}{5}\)
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{10}\)
...
\(\dfrac{1}{10}< \dfrac{1}{5}\)
Do đó: \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{10}< \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{5}=\dfrac{6}{5}\)
Ta có: \(\dfrac{1}{11}=\dfrac{1}{11}\)
\(\dfrac{1}{12}< \dfrac{1}{11}\)
...
\(\dfrac{1}{17}< \dfrac{1}{11}\)
Do đó: \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{17}< \dfrac{1}{11}+\dfrac{1}{11}+...+\dfrac{1}{11}=\dfrac{7}{11}\)
Vì \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{10}< \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{5}=\dfrac{6}{5}\)
và \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{17}< \dfrac{1}{11}+\dfrac{1}{11}+...+\dfrac{1}{11}=\dfrac{7}{11}\)
nên \(P< \dfrac{6}{5}+\dfrac{7}{11}=\dfrac{101}{55}< \dfrac{110}{55}=2\)
hay P<2
Mình làm lại nha, nãy bị lỗi tí!!
Ta thấy:
\(\dfrac{1}{6}\)<\(\dfrac{1}{5}\)
\(\dfrac{1}{7}\)<\(\dfrac{1}{5}\)
\(\dfrac{1}{8}\)<\(\dfrac{1}{5}\)
.........
\(\dfrac{1}{17}< \dfrac{1}{10}\)
=>\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+....+\dfrac{1}{17}< 5.\dfrac{1}{5}+8.\dfrac{1}{10}=1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
Vậy P<2
P=1/5+1/6+1/7+...+1/17
P=(1/17+1/5)[(1/17-1/5)+1]:2=11/73
\(\Rightarrow\)P<2
