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We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$ Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.

Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$ for all $x>e$.

Thus, for all $n\geq3$ we have $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ Now,for $n=2$ we get $$\sqrt2<\sqrt[3]3,$$ and for $n=1$ we have $1^1<2^\frac{1}{2}$, which gives that $\sqrt[3]3$ is a largest number in the sequence.

Done!

We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$ Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.

Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$ for all $x>e$.

Thus, for all $n\geq3$ we have $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ Now, for $n=2$ we get $$\sqrt2<\sqrt[3]3,$$ and for $n=1$ we have $1^1<2^\frac{1}{2}$, which gives that $\sqrt[3]3$ is a largest number in the sequence.

Done!

We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$ Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.

Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$ for all $x>e$.

Thus, for all $n\geq3$ we have $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ Now,for $n=2$ need to prove that $$\sqrt2>\sqrt[3]3,$$ which is wrong and easy to see that your statement is wrong for $n=1$.

Done!

We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$ Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.

Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$ for all $x>e$.

Thus, for all $n\geq3$ we have $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ Now,for $n=2$ we get $$\sqrt2<\sqrt[3]3,$$ and for $n=1$ we have $1^1<2^\frac{1}{2}$, which gives that $\sqrt[3]3$ is a largest number in the sequence.

Done!

We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$

We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$ Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.

Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$ for all $x>e$.

Thus, for all $n\geq3$ we have $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ Now,for $n=2$ need to prove that $$\sqrt2>\sqrt[3]3,$$ which is wrong and easy to see that your statement is wrong for $n=1$.

Done!