Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{S_n} = \frac{{{1^2} - 1}}{1} + \frac{{{2^2} - 1}}{{{2^2}}} + \frac{{{3^2} - 1}}{{{3^2}}} + ..... + \frac{{{n^2} - 1}}{{{n^2}}}\\
= 1 - \frac{1}{1} + 1 - \frac{1}{{{2^2}}} + 1 - \frac{1}{{{3^2}}} + .... + 1 - \frac{1}{{{n^2}}}\\
= n - \left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{n^2}}}} \right)\\
A = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{n^2}}}\\
< 1 + \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + .... + \frac{1}{{\left( {n - 1} \right).n}}\\
= 1 + 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + .... + \frac{1}{{n - 1}} - \frac{1}{n}\\
= 2 - \frac{1}{n} < 2\\
A = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{n^2}}} > 1\\
\Rightarrow {S_n} = n - A \Rightarrow n - 2 < {S_n} < n - 1
\end{array}\)
Suy ra Sn không là số nguyên.
