TITLE: Find the largest term of the sequence $a_n=\sqrt[n]{n}$
QUESTION [8 upvotes]: Find the largest term of the sequence $a_n=\sqrt[n]{n}$. 

By simple calculation:
$$a_1= 1$$
$$a_2=1.41$$
$$a_3=1.44$$
$$a_4=1.41$$
$$a_5=1.37$$
$$a_6=1.348$$
$$\quad\vdots$$
After that the sequence seems to be pretty much decreasing and 
$$\lim_{n\to \infty}{\sqrt[n]{n}}=1$$
This way it looks like $a_3$ is the largest term however there is no official proof behind this. 
What's the usual way to approach such problems?

REPLY [17 votes]: You can use the extension to the real line and find the maximum by differentiation (of the logarithm, for convenience):
$$\left(\frac{\log x}x\right)'=\frac{1-\log x}{x^2}=0$$
Hence the function is decreasing on either sides of $x=e$ and the maximum for the discrete variable is one of $a_2, a_3$.