Có S=\(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
=>\(\dfrac{1}{2^2}S=\dfrac{1}{2^2}\)\(\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
=> \(\dfrac{1}{2^2}\)S= \(\dfrac{1}{2^4}-\dfrac{1}{2^6}+\dfrac{1}{2^8}-...+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+...+\dfrac{1}{2^{2004}}-\dfrac{1}{2^{2006}}\)
+S =\(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
=> \(\dfrac{5}{4}\)S= \(\dfrac{1}{2^2}\)-\(\dfrac{1}{2^{2006}}\)
=> S= \(\dfrac{\left(\dfrac{1}{2^2}-\dfrac{1}{2^{2006}}\right)}{\dfrac{5}{2^2}}=\dfrac{\dfrac{1}{2^2}}{\dfrac{5}{2^2}}-\dfrac{\dfrac{1}{2^{2006}}}{\dfrac{5}{2^2}}=\dfrac{1}{5}-\dfrac{1}{2^{2004}.5}=0.2-\dfrac{1}{2^{2004}.5}\)
=> S <0,2
Vậy S <0,2(đpc/m)