Question

Cho S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{18^2}\)+\(\frac{1}{19^2}\). So sánh S với \(^{\frac{18}{19}}\)

Answer 1

S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{18^2}\)+\(\frac{1}{19^2}\)

S<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{17.18}\)+\(\frac{1}{18.19}\)

S<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{17}\)-\(\frac{1}{18}\)+\(\frac{1}{18}\)-\(\frac{1}{19}\)

S<1-\(\frac{1}{19}\)

\(\Rightarrow\)S<\(\frac{18}{18}\)

Answer 2

Mk viết nhầm nhé,S<\(\frac{18}{19}\)

Answer 3

Có: \(\frac{1}{2^2}< \frac{1}{1\cdot2}\\ \frac{1}{3^2}< \frac{1}{2\cdot3}\\ \frac{1}{4^2}< \frac{1}{3\cdot4}\\ \frac{1}{5^2}< \frac{1}{4\cdot5}\\...\\ \frac{1}{18^2}< \frac{1}{17\cdot18}\\ \frac{1}{19^2}< \frac{1}{18\cdot19}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{18^2}+\frac{1}{19^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{17\cdot18}+\frac{1}{18\cdot19}\\ \Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{18}+\frac{1}{18}-\frac{1}{19}\\ \Rightarrow S< 1-\frac{1}{19}\\ \Rightarrow S< \frac{18}{19}\)

Vậy \(S< \frac{18}{19}\)