CMR $\frac{1}{\sqrt{k}(k+1)}<$ $2(\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1}}$ 01/07/2021 Bởi Eva CMR $\frac{1}{\sqrt{k}(k+1)}<$ $2(\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1}}$
Bạn tham khảo: $\frac{\sqrt{k}}{k(k+1)}=$ $\sqrt{k}$.($\frac{1}{k}-$ $\frac{1}{k+1})$ =$\sqrt{k} $.($\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1}}).($ $\frac{1}{\sqrt{k}}+$ $\frac{1}{\sqrt{k+1}})$ =($1+\frac{\sqrt{k}}{\sqrt{k+1}})($ $\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1})}<2$($\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1}})$ Bình luận
Bạn tham khảo:
$\frac{\sqrt{k}}{k(k+1)}=$ $\sqrt{k}$.($\frac{1}{k}-$ $\frac{1}{k+1})$
=$\sqrt{k} $.($\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1}}).($ $\frac{1}{\sqrt{k}}+$ $\frac{1}{\sqrt{k+1}})$
=($1+\frac{\sqrt{k}}{\sqrt{k+1}})($ $\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1})}<2$($\frac{1}{\sqrt{k}}-$ $\frac{1}{\sqrt{k+1}})$