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?
Best Answer
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 $a_2, a_3$.