Đáp án:

$\begin{array}{l}
S = \dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{1}{{36}} + \dfrac{1}{{64}} + \dfrac{1}{{100}} + \dfrac{1}{{144}} + \dfrac{1}{{196}}\\
 = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}} + \dfrac{1}{{{8^2}}} + \dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{12}^2}}} + \dfrac{1}{{{{14}^2}}}\\
 = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^2}{{.2}^2}}} + \dfrac{1}{{{2^2}{{.3}^2}}} + \dfrac{1}{{{2^2}{{.4}^2}}} + \dfrac{1}{{{2^2}{{.5}^2}}} + \dfrac{1}{{{2^2}{{.6}^2}}} + \dfrac{1}{{{2^2}{{.7}^2}}}\\
 = \dfrac{1}{{{2^2}}}.\left( {1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + ... + \dfrac{1}{{{7^2}}}} \right)\\
 < \dfrac{1}{{{2^2}}}\left( {1 + \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{6.7}}} \right)\\
 \Leftrightarrow S < \dfrac{1}{4}.\left( {1 + 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{6} - \dfrac{1}{7}} \right)\\
 \Leftrightarrow S < \dfrac{1}{4}.\left( {2 - \dfrac{1}{7}} \right)\\
 \Leftrightarrow S < \dfrac{1}{2} - \dfrac{1}{{28}} < \dfrac{1}{2}\\
Vậy\,S < \dfrac{1}{2}
\end{array}$

1/4 = 1/(22) < 1/(12) = 1/2 - 1/4 

tương tự ta có 1/16 < 1/(24) = 1/4 - 1/8 1/36 < 1/(46) = 1/8 - 1/12 1/64 < 1/(68) = 1/12 - 1/16 1/100 < 1/(810) = 1/16 - 1/20 1/144 < 1/(1012) = 1/20 - 1/24 1/196 < 1/(12 14) = 1/24 - 1/28 cộng hết lại => 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm