An AM-GM inequality for the harmonic numbers If $H_n=1+\frac 12+\frac 13+...+\frac 1n(n>2)$, prove that $$n(n+1)^{1/n}-n<H_n<n-(n-1)n^{-\frac 1{n-1}}$$ I did the following $$\frac {1+\frac 12+\frac 13+...+\frac 1n}n\ge \frac 1{(n!)^{1/n}}\\=>H_n\ge n\left(\frac 1{(n!)^{1/n}}\right)$$ Hence we have to prove for the left inequality that $$\begin{align}&n\left(\frac 1{(n!)^{1/n}}\right)\ge n[(n+1)^{1/n}-1]\\&=>1\ge [(n+1)!]^{1/n}-[n!]^{1/n}\end{align}$$ How do I proceed after this? Also I have no clue about how to approach the right inequality. 

We have:
$$ H_n + n =\sum_{k=1}^{n}\left(1+\frac{1}{k}\right) $$
hence, by the AM-GM inequality,
$$ H_n + n \geq n\,\left(\prod_{k=1}^{n}\frac{k+1}{k}\right)^{1/n}=n(n+1)^{1/n}.$$
On the other hand,
$$ n-H_n = \sum_{k=2}^{n}\left(1-\frac{1}{k}\right),$$
hence, by the AM-GM inequality again:
$$ n-H_n \geq (n-1)\,\left(\prod_{k=2}^{n}\frac{k-1}{k}\right)^{1/(n-1)}=(n-1)\, n^{-1/(n-1)}. $$
