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?
On
Hint $$a_n \leq a_{n+1} \Leftrightarrow n^{n+1} \leq (n+1)^n \Leftrightarrow n \leq (1+\frac{1}{n})^n$$
Now use the fact that $(1+\frac{1}{n})^n$ is increasing to $e$.
On
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 $a_2, a_3$.
$$\sqrt[n]n>\sqrt[n+1]{n+1}$$ it's $$n>\left(1+\frac{1}{n}\right)^n,$$ which is obvious for $n\geq3$ because $$\left(1+\frac{1}{n}\right)^n<e<3.$$ Thus, by your work for $n\leq2$ we see that $a_3$ is a maximum.