For every positive integer $n$ set $H_n = \dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$. Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.


I will prove this result by induction. The base case holds since for $n = 2$ we have $(1+2)^{\frac{1}{2}} < 2 + H_2 = 3.5$. Now assume that $k(1+k)^{\frac{1}{k}} < k+H_k$ holds for some $k$. We need to show that $(k+1)(2+k)^{\frac{1}{k+1}} < 1+k+H_{k+1}$. How do I show this? Also, is induction the best way to prove this?

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    $\begingroup$ I do not think induction would be a good idea as the 2 expressions you obtained do not look close to each other. $\endgroup$ – Element118 Dec 28 '15 at 23:25
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    $\begingroup$ @ReneSchipperus It is $\left(1+\frac{1}{n}\right)^n$ that approaches $e$. $\endgroup$ – Element118 Dec 28 '15 at 23:36

Notice that $$n+1 = \frac{2}{1}\frac{3}{2} \cdots \frac{n}{n-1}\frac{n+1}{n} $$ Hence by AM-GM we get $$ \left (n+1\right )^{\frac{1}{n}} < \frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n} }{n} $$ But observe that $$\frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n} }{n}= \frac{n + H_n}{n}.$$ And we are done.

  • $\begingroup$ Nice. Can you show $n(\sqrt[n]{n+1}-1)\leq \ln n$ ? $\endgroup$ – Rene Schipperus Dec 28 '15 at 23:55
  • $\begingroup$ You might want to make that inequality sign a strict inequality, as in the question. $\endgroup$ – Element118 Dec 28 '15 at 23:58
  • $\begingroup$ @Element118 thanks for catching that $\endgroup$ – clark Dec 29 '15 at 0:00
  • $\begingroup$ @ReneSchipperus thanks, not right off the bat, I will think about it. $\endgroup$ – clark Dec 29 '15 at 0:02
  • $\begingroup$ @Rene It fails at least for $n=1$. $\endgroup$ – A.S. Dec 29 '15 at 0:04

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