# Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.

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 $n \geq 2$.

Attempt

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?

• I do not think induction would be a good idea as the 2 expressions you obtained do not look close to each other. – Element118 Dec 28 '15 at 23:25
• @ReneSchipperus It is $\left(1+\frac{1}{n}\right)^n$ that approaches $e$. – 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.
• Nice. Can you show $n(\sqrt[n]{n+1}-1)\leq \ln n$ ? – Rene Schipperus Dec 28 '15 at 23:55
• @Rene It fails at least for $n=1$. – A.S. Dec 29 '15 at 0:04