$\frac{1}{4^{2} }$ +$\frac{1}{6^{2}}$+ $\frac{1}{8^{2}}$+...+ $\frac{1}{(2n)^{2}}$
=$\frac{1}{16}$+ $\frac{1}{36}$+ $\frac{1}{64}$+...+$\frac{1}{4.n^{2}}$
=$\frac{1}{4}$( $\frac{1}{4}$+ $\frac{1}{9}$+ $\frac{1}{16}$+...+ $\frac{1}{n^{2}}$)
=$\frac{1}{4}$( $\frac{1}{2^{2}}$+ $\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+...+ $\frac{1}{n^{2}}$)
Ta có:
$\frac{1}{k^{2}}$<$\frac{1}{k.(k-1)}$
Do đó nên ta có:
$\frac{1}{2^{2}}$< $\frac{1}{1.2}$
$\frac{1}{3^{2}}$< $\frac{1}{2.3}$
$\frac{1}{4^{2}}$< $\frac{1}{3.4}$
..............
$\frac{1}{n^{2}}$< $\frac{1}{n.(n-1)}$
⇒$\frac{1}{4}$( $\frac{1}{2^{2}}$+ $\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+...+ $\frac{1}{n^{2}}$) <$\frac{1}{4}$( $\frac{1}{1.2}$+$\frac{1}{2.3}$+ $\frac{1}{3.4}$+...+ $\frac{1}{(n-1).n}$) =$\frac{1}{4}$(1- $\frac{1}{2}$+ $\frac{1}{2}$- $\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+...+ $\frac{1}{n-1}$- $\frac{1}{n}$) =$\frac{1}{4}$(1- $\frac{1}{n}$)=$\frac{1}{4}$- $\frac{1}{4n}$ <$\frac{1}{4}$
Vậy $\frac{1}{4^{2} }$ +$\frac{1}{6^{2}}$+ $\frac{1}{8^{2}}$+...+ $\frac{1}{(2n)^{2}}$<$\frac{1}{4}$
CHÚC BẠN HỌC TỐT!!!
