Find the maximum among \mathbf { 1 , 2^{1/2} , 3^{1/3} , 4^{1/4} , ... } .

Discussion

Consider the function \mathbf { f(x) = x^{1/x} } . We employ standard techniques to compute the maxima.

Take logarithm on both sides we have \mathbf { \log f(x) = \frac{1}{x} \log x } . Next find out the derivative:

\mathbf {\frac {1}{f(x)} f'(x) = \frac{-1}{x^2} \log x + \frac{1}{x}\cdot\frac{1}{x} implies f'(x) = f(x) \cdot \frac{1}{x^2} (1 - \log x) }

Since \mathbf { f(x) = x^{1/x} } is always positive for positive x and so is \mathbf {\frac{1}{x^2}} sign of the derivative depends only on (1-logx). Hence the derivative is 0 at x = e (2.71 approximately), positive before that and negative after that. Hence the function has a maxima at x = e.

We check the values at x=2 and x=3 and easy computations show that \mathbf { 3^{1/3} > 2^{1/2} }. Hence \mathbf {3^{1/3} } is the largest value.

Special Note

One may ask for a non calculus proof of this problem. The basic idea is to understand that the inequality
\mathbf { n^{1/n} > (n+1)^{1/n+1}\Rightarrow n^{n+1} > (n+1)^n \Rightarrow n\cdot n^n > (n+1)^n \\ \Rightarrow n > \frac{(n+1)^n}{n^n}\Rightarrow n > (1+ \frac{1}{n})^n }

It is easy to show that the quantity \mathbf { (1+ \frac{1}{n})^n } lies within 2 and 3 for all values of n. Hence the inequality \mathbf {n > (1+ \frac{1}{n})^n } is true for n > 3. The result follows.