Đáp án:
Giải thích các bước giải:
Ta có
$\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + ... + \frac{1}{10000} $
$= \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + ... + \frac{1}{100^2} $
$= \frac{1}{4} ( 1 + \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{50^2} ) $
$< \frac{1}{4} ( 1 + \frac{1}{1.2) + \frac{1}{2.3) + \frac{1}{3.4} + ... + \frac{1}{49.50} ) $
$=\frac{1}{4}(1+ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{49 } - \frac{1}{50} $
$= \frac{1}{4}( 2 - \frac{1}{50} ) $
$= \frac{1}{4} . \frac{99}{50} $
$= \frac{99}{200} < \frac{1}{2} $
Vậy $\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + ... + \frac{1}{10000} < \frac{1}{2} $
