\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1^{ }}{4^2}\) + ....... + \(\frac{1}{100^2}\)= \(\frac{1}{2x2}\) + \(\frac{1}{3x3}\) + \(\frac{1}{4x4}\) + .... + \(\frac{1}{100x100}\)
Mà \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + .... + \(\frac{1}{100x100}\) < \(\frac{1}{1x2}\) + \(\frac{1}{2x3}\) +....+ \(\frac{1}{99x100}\)
=> \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + .... + \(\frac{1}{100.100}\) < 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .... + \(\frac{1}{99}\) - \(\frac{1}{100}\)
=> \(\frac{1}{2.2}\) + \(\frac{1}{3x3}\) + .... + \(\frac{1}{100x100}.\) < 1 - \(\frac{1}{100}\) < 1
=> \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + .... + \(\frac{1}{100^2}\) < 1
k ủng hộ mình nhaaaaa?!!!!
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}< 1\)\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{100^2}< 1\)
