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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?

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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 $a_2, a_3$.

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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 $e$.

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

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  • $\begingroup$ 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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