difficult sequence Prove that the sequence:
$$a_n= \frac{1}{n}\left(e\cdot\sqrt e \cdots\sqrt[3]e\cdot\sqrt[n]e\right)$$
is decreasing to a finite limit. After having shown that the sequence: 
$$b_n=\left(\sum_{k=1}^n\frac{1}{k}\right)-\log n$$
converges to a positive real number $b,$ say  who  is the limit of $ a_n $

$$\eqalign{
  & {a_n} = \frac{1}{n}\prod\limits_{k = 1}^n {{e^{\frac{1}{k}}}}   \cr 
  & \log {a_n} =  - \log n + \sum\limits_{k = 1}^n {\frac{1}{k}}   \cr 
  & \log {a_n} = \sum\limits_{k = 1}^n {\frac{1}{k}}  - \log n  \cr 
  & \mathop {\lim }\limits_{n \to \infty } \log {a_n} = \gamma  \cr} $$
You can prove $0 < \gamma < 1$ since we can replace $\log n$ by $\log (n+1)$ and put
$$\eqalign{
  & {b_n} = \sum\limits_{k = 1}^n {\frac{1}{k}}  - \sum\limits_{k = 1}^n {\log \frac{{k + 1}}{k}}   \cr 
  & {b_n} = \sum\limits_{k = 1}^n {\left( {\frac{1}{k} - \log \frac{{k + 1}}{k}} \right)}   \cr} $$
Since we know 
$$1 - \frac{1}{x} \leqslant \log x \leqslant x - 1$$
We have
$$\frac{1}{{k + 1}} \leq \log \left( {1 + \frac{1}{k}} \right) \leq \frac{1}{k}$$
We can prove both bounds with this. 
$$\eqalign{
  & \frac{1}{k} - \frac{1}{{k + 1}} \geqslant \frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0  \cr 
  & \sum\limits_{k = 1}^n {\frac{1}{k} - \frac{1}{{k + 1}}}  \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)}  \geqslant 0  \cr 
  & 1 \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)}  \geqslant 0 \cr} $$
The first inequality also proves each term is positive or zero (though the last is discarded by a simple look at the image), i.e
$$\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0$$  

