Do S = \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}\)

\(\Rightarrow\)S = \(\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)

\(\Rightarrow\)S=(1+1+1+...+1) - \(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(\Rightarrow\)S=49-\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Dễ thấy:\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)không phải là số tự nhiên

\(\Rightarrow\)S\(\notin N\)

Ta có : \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{15}+...+\frac{1}{10000}\right)\)

\(=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99\)

\(\Rightarrow\)S<99 (1)

Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)

\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

...

\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< 1-\frac{1}{100}< 1\)

\(\Rightarrow\)S>99-1=98 (2)

Từ (1) và (2)

\(\Rightarrow\)98<S<99

\(\Rightarrow\)S\(\notin\)N

Vậy S\(\notin\)N.