Điền dấu >;<; =; vào ô trống:

Ta có:

\(\frac{1}{{31}} > \frac{1}{{60}}\);     \(\frac{1}{{32}} > \frac{1}{{60}}\);    ...;      \(\frac{1}{{59}} > \frac{1}{{60}}\);           \(\frac{1}{{60}} = \frac{1}{{60}}\).

Nên: \(\frac{1}{{31}} + \frac{1}{{32}} + \frac{1}{{33}} + ... + \frac{1}{{60}} > \underbrace {\frac{1}{{60}} + \frac{1}{{60}} + \frac{1}{{60}}}_{} + ... + \frac{1}{{60}} = 30 \times \frac{1}{{60}} = \frac{1}{2}\)

30 phân số \(\frac{1}{{60}}\)

Tương tự ta có:

\(\frac{1}{{61}} > \frac{1}{{90}}\);     \(\frac{1}{{62}} > \frac{1}{{90}}\);    ...;      \(\frac{1}{{89}} > \frac{1}{{90}}\); \(\frac{1}{{90}} = \frac{1}{{90}}\).

Nên: \(\frac{1}{{61}} + \frac{1}{{62}} + ... + \frac{1}{{89}} + \frac{1}{{90}} > \underbrace {\frac{1}{{90}} + \frac{1}{{90}} + ... + \frac{1}{{90}} + \frac{1}{{90}}}_{} = 30 \times \frac{1}{{90}} = \frac{1}{3}\)

30 phân số \(\frac{1}{{90}}\)

Do đó: \(\frac{1}{{31}} + \frac{1}{{32}} + \frac{1}{{33}} + ... + \frac{1}{{89}} + \frac{1}{{90}} > \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\)