Let $k : \mathbb{N}^{>0} \to \mathbb{R}$ be such that $k_n = n(t^{1/n} - 1)$, where $t \in [0, 1] \subset \mathbb{R}$. Note that $\lim_{n \to \infty} k_n = \log(t)$. The plot of $k$ for a given $t$ seems to be decreasing. Prove that $k$ is decreasing.

  • $\begingroup$ I would have thought $n(t^{1/n} - 1) \le 0$ if $0 \le t \le 1$ $\endgroup$ – Henry Nov 17 '14 at 21:52
  • $\begingroup$ Yes, and $\log(t)\lt0$ for $t$ in $(0,1)$... $\endgroup$ – Did Nov 17 '14 at 22:00

Let $s=t^{1/(n(n+1))}$ then $t^{1/n}=s^{n+1}$ and $t^{1/(n+1)}=s^n$ hence $k_{n+1}-k_n=u(s)$ where, for every $x$, $u(x)=(n+1)x^n-nx^{n+1}-1$. One sees that $u(1)=0$ and that $u'(x)=n(n+1)x^{n-1}(1-x)\gt0$ for every $x$ in $(0,1)$. This implies that $u(x)\lt0$ for every $x$ in $(0,1)$ hence $k_{n+1}\lt k_n$.

  • $\begingroup$ That's clever, and seems valid to me. I had trouble because of the differing exponents of t in the forward-difference. This takes care of them neatly right at the beginning. $\endgroup$ – kaba Nov 17 '14 at 22:17
  • $\begingroup$ Yes, the trick is not necessary for the proof, but it might make it easier to follow. $\endgroup$ – Did Nov 17 '14 at 22:25

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