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[TEX]\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{(2n-1)(2n+1)}=2(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2n-1}-\frac{1}{2n+1})=2(1-\frac{1}{2n+1})=\frac{4n}{2n+1}\\ \Rightarrow \lim \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{(2n-1)(2n+1)}=\lim \frac{4n}{2n+1}=2[/TEX]

[TEX]1+2+...+n=\frac{n^2+n}{2}\\ \Rightarrow\lim \frac{1+2+...+n}{n^2+3n}=\lim\frac{n^2+n}{2(n^2+3n)}=\frac{1}{2}[/TEX]
 
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