Solve: $$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{1+\sqrt[3]{3}+...\sqrt[2n-1]{2n-1}})^n$$
Solve:$$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{1+\sqrt[3]{3}+...\sqrt[2n-1]{2n-1}})^n$$
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Solve: $$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{
\sqrt[3]{1
}+\sqrt[3]{3}+...\sqrt[2n-1]{2n-1}})^n$$
Solve:$$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{
\sqrt[3]{1
}+\sqrt[3]{3}+...\sqrt[2n-1]{2n-1}})^n$$
Một số quy tắc tìm giới...
Solve: $$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{1+\sqrt[3]{3}+...\sqrt[2n-1]{2n-1}})^n$$
Solve:$$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{1+\sqrt[3]{3}+...\sqrt[2n-1]{2n-1}})^n$$
Một số quy tắc tìm giới...