Áp dụng công thức \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\). Ta có:
\(\frac{1}{2^2}< 2-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
. . . . .
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)
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\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2-\frac{1}{50}=\frac{99}{50}\)
Vậy:A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2^{\left(đpcm\right)}\)