$P=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+....+ \frac{1}{n^{2}}<\frac{5}{3} (n\epsilon N, n\geq1)$

$$P=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+....+ \frac{1}{n^{2}}<\frac{5}{3} (n\epsilon N, n\geq1)$$
Với mọi $k\geq 1$ ta có
$$\frac{1}{k^2}=\frac{4}{4k^2}<\frac{4}{4k^2-1}=2(\frac{1}{2k-1}-\frac{1}{2k+1})$$
Cho $k=2,3,4,...,n$ ta có
$$\frac{1}{2^2}=\frac{4}{4.2^2}<\frac{4}{4.2^2-1}=\frac{2}{2.2-1}-\frac{2}{2.2+1}=\frac{2}{3}-\frac{2}{5}$$
$$\frac{1}{3^2}=\frac{4}{4.3^2}<\frac{4}{4.3^2-1}=\frac{2}{2.3-1}-\frac{2}{2.3+1}=\frac{2}{5}-\frac{2}{7}$$
..............................................
$$\frac{1}{n^2}=\frac{4}{4n^2}<\frac{4}{4n^2-1}=\frac{2}{2n-1}-\frac{2}{2n+1}$$
Cộng lài ta được$VT<1+\frac{2}{3}-\frac{2}{2n+1}<1+\frac{2}{3}=\frac{5}{3}$
Đpcm $\blacksquare$