Hướng Dẫn Giải

Ta có: \(\frac{1}{4} + \frac{1}{9} + \frac{1}{{16}} + \frac{1}{{25}} + ... + \frac{1}{{9801}} + \frac{1}{{10000}}\)

\( = \frac{1}{4} + \frac{1}{{3 \times 3}} + \frac{1}{{4 \times 4}} + \frac{1}{{5 \times 5}} + ... + \frac{1}{{99 \times 99}} + \frac{1}{{100 \times 100}}\)

Thấy: \(\frac{1}{4} + \left( {\frac{1}{{3 \times 3}} + \frac{1}{{4 \times 4}} + \frac{1}{{5 \times 5}} + ... + \frac{1}{{99 \times 99}} + \frac{1}{{100 \times 100}}} \right)\)

\( < \frac{1}{4} + (\frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + ... + \frac{1}{{98 \times 99}} + \frac{1}{{99 \times 100}})\)

Ta có: \(\frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + ... + \frac{1}{{98 \times 99}} + \frac{1}{{99 \times 100}}\)

\( = \frac{{3 - 2}}{{2 \times 3}} + \frac{{4 - 3}}{{3 \times 4}} + ... + \frac{{99 - 98}}{{98 \times 99}} + \frac{{100 - 99}}{{99 \times 100}}\)

\( = \frac{3}{{2 \times 3}} - \,\frac{2}{{2 \times 3}} + \frac{4}{{3 \times 4}} - \frac{4}{{3 \times 4}} + ... + \frac{{99}}{{98 \times 99}} - \frac{{98}}{{98 \times 99}} + \frac{{100}}{{99 \times 100}} - \frac{{99}}{{99 \times 100}}\)

\( = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{98}} - \frac{1}{{99}} + \frac{1}{{99}} - \frac{1}{{100}}\)\( = \frac{1}{2} - \frac{1}{{100}}\)

Suy ra \(\frac{1}{4} + \frac{1}{9} + \frac{1}{{16}} + \frac{1}{{25}} + ... + \frac{1}{{9801}} + \frac{1}{{10000}} < \frac{1}{4} + \frac{1}{2} - \frac{1}{{100}} = \frac{3}{4} - \frac{1}{{100}} < \frac{3}{4}\)

Vậy \(\frac{1}{4} + \frac{1}{9} + \frac{1}{{16}} + \frac{1}{{25}} + ... + \frac{1}{{9801}} + \frac{1}{{10000}} < \frac{3}{4}\)

Trong các phân số sau đây, phân số nào lớn hơn 1: