Question

[Vũ Thị Vân Anh]

Cho \(S=\dfrac{1}{5^2}+\dfrac{1}{9^2}+....................+\dfrac{1}{409^2}\) Chứng minh rằng \(S< \dfrac{1}{12}\)

Answer 1

[Akai Haruma]

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.

Answer 2

[hải yến phạm]

hay