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Show that $$ a_n=n^{1\over n}$$ is eventually decreasing.

I am not allowed to used derivatives and I have been trying for a while. I graphed it and it is decreasing if $n>2$, but how can you show it?

I have tried to look at ${a_{n+1}/a_n}$ but it gets me nowhere.

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    $\begingroup$ $$n^{\frac{1}{n}} > (n+1)^{\frac{1}{n+1}} \iff n^{n+1} > (n+1)^n$$ $\endgroup$
    – Daniel Fischer
    Oct 12, 2015 at 18:56

3 Answers 3

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Assume $(n-1)^n\gt n^{n-1}$, which is easily verified for $n=4$. Then

$$n^{n+1} =n\left(n\over(n-1)\right)^n(n-1)^n \gt \left({n^2\over n-1}\right)^n =\left({n^2-1+1\over n-1}\right)^n =\left(n+1+{1\over n-1}\right)^n \gt(n+1)^n$$

The first inequality uses the induction hypothesis; the second uses the fact that $n\gt1$.

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    $\begingroup$ You have an extra $n$ that should not be there after the first inequality sign. $\endgroup$
    – Vincent Pfenninger
    Oct 12, 2015 at 20:28
  • $\begingroup$ @VincentPfenninger, are you referring to the initial $n$, before the term that's raised to the $n$th power? If so, I left it there intentionally; its removal gives the second inequality. $\endgroup$
    – Barry Cipra
    Oct 12, 2015 at 20:32
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    $\begingroup$ After using the induction hypothesis you are left with $n\left(\frac{n}{n-1}\right)^nn^{n-1} = \left(\frac{n}{n-1}\right)^nn^{n} = \left(\frac{n^2}{n-1}\right)^n$. $\endgroup$
    – Vincent Pfenninger
    Oct 12, 2015 at 20:36
  • $\begingroup$ Oh gosh, you're quite right. I'll fix it. Thank you! $\endgroup$
    – Barry Cipra
    Oct 12, 2015 at 20:43
  • $\begingroup$ This proof doesn't seem right to me. The usage of induction proves this only for integer values of n $\endgroup$
    – JobHunter69
    Oct 22, 2023 at 3:40
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Show that if $n\geq 3$ then $$\left(1+\frac{1}{n}\right)^n<n.$$

You can actually show by elementary means (without logarithms, calculus, or $e$, just binomial theorem and some simple inequalities) that:

$$\left(1+\frac{1}{n}\right)^n\leq 3$$

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The solution follows Daniel Fischer's hint.

Using the inequality (for $x \ne 0$) $$ 1 + x < e^x, $$ with $x = 1/n$, we have $$ \frac{n+1}{n} \le e^{1/n}. $$ But for $n \ge 3 > e$ $$ e^{1/n} < n^{1/n}. $$ So $$ \frac{n+1}{n} < n^{1/n}, $$ or $$ n+1 < n^{\frac{n+1}{n}}, $$ Taking the $(n+1)$th root yields the desired result.

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  • $\begingroup$ Oh that was a great help from you $\endgroup$
    – Umesh shankar
    Jun 19, 2021 at 12:50

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