Find the largest term of the sequence $a_n=\sqrt[n]{n}$.

By simple calculation:

$$a_1= 1$$







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?


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 $a_2, a_3$.


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 $e$.


$$\sqrt[n]n>\sqrt[n+1]{n+1}$$ 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.

  • $\begingroup$ The inequations are wrong for $n=1, 2$. Had the square root of $2$ been $1.5$, the conclusion would have been wrong. You should add $a_1,a_2<a_3$. $\endgroup$ – Yves Daoust Jan 26 '18 at 8:23
  • $\begingroup$ See please better my post. I said about $n\geq3$. About $n\leq2$ see the starting post. $\endgroup$ – Michael Rozenberg Jan 26 '18 at 10:09
  • $\begingroup$ IMO, that should be said explicitly here. $\endgroup$ – Yves Daoust Jan 26 '18 at 11:23

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