Question

[Dương]

chứng minh: 499/1000<1/2^2+1/3^2+1/4^2+...+1/999^2<3/4

Answer 1

[✎﹏ミ★꧁༺вєѕт↭ℓαυяιєℓ↭νи༻꧂★ミ.༻(Trưở...]

ta có:

\(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};.....;\frac{1}{999^2}>\frac{1}{999.1000}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{999^2}>\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{999.1000}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{999^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{999}-\frac{1}{1000}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{999^2}>\frac{1}{2}-\frac{1}{1000}=\frac{499}{1000}\)