Lời giải:
Ta có:
\(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{3.7}\)
\(\frac{1}{9^2}=\frac{1}{9.9}< \frac{1}{7.11}\)
.......
\(\frac{1}{409^2}=\frac{1}{409.409}=\frac{1}{(407+2)(411-2)}=\frac{1}{407.411-2.407+2.411}< \frac{1}{407.411}\)
Cộng theo vế ta có:
\(S<\frac{1}{3.7}+\frac{1}{7.11}+....+\frac{1}{407.411}(*)\)
Mà:
\(\frac{1}{3.7}+\frac{1}{7.11}+....+\frac{1}{407.411}=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{407.411}\right)\)
\(=\frac{1}{4}\left(\frac{7-3}{3.7}+\frac{11-7}{7.11}+....+\frac{411-407}{407.411}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{407}-\frac{1}{411}\right)\)
\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{411}\right)< \frac{1}{4}.\frac{1}{3}=\frac{1}{12}(**)\)
Từ \((*); (**)\Rightarrow S< \frac{1}{12}\)
Ta có đpcm.
