\(S=\frac{5}{2^2}+\frac{5}{3^2}+\frac{5}{4^2}+...+\frac{5}{100^2}\)
\(S=5.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Ta có : \(\frac{1}{2^2}>\frac{1}{2.3},\frac{1}{3^2}>\frac{1}{3.4},\frac{1}{4^2}>\frac{1}{4.5},...,\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow5.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)>5.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(\Rightarrow S>5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow S>5.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow S>5.\frac{99}{202}\)
\(\Rightarrow S>\frac{495}{202}>\frac{404}{202}=2\)
\(\Rightarrow S>2\)
\(CM:S< 5\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2},\frac{1}{3^2}< \frac{1}{2.3},...,\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}\)
\(\Rightarrow5.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)< 5.\frac{99}{100}\)
\(\Rightarrow S< \frac{495}{100}< \frac{500}{100}\)
\(\Rightarrow S< 5\)
