<s>[INST] Need help proving an interval $\frac {1} {ek} \le \frac {1}{k} (1 - \frac {1}{k} )^{k-1} \le \frac {1}{2k}$ I am trying to proof
$$\frac {1} {ek}  \le  \frac {1}{k}   (1 - \frac {1}{k} )^{k-1} \le \frac {1}{2k} $$ 
for k>=2
to prove this I first multiply by k getting
$$\frac {1} {e}  \le  \left(1 - \frac {1}{k} \right)^{k-1} \le \frac {1}{2} $$ 
then use case $k=2$ as a base case
$$\frac {1} {e}  <=  \frac {1}{2}  <= \frac {1}{2} $$ 
which is good, then assumed 
$$\frac {1} {e}  <=  (1 - \frac {1}{k} )^{k-1}  <= \frac {1}{2} $$  
to be true for any k>2 and try to prove for k+1
so I sustitute k+1 on k getting
$$\frac {1} {e}  <=  (1 - \frac {1}{k+1} )^{k}  <= \frac {1}{2} $$ 
which equals
$$\frac {1} {e}  <=  (\frac {k}{k+1} )^{k}  <= \frac {1}{2} $$ 
so I am trying to get $$ (\frac {k}{k+1} )^{k}  $$ to any of the original formulas to finish the prove but I have been unsuccesful. I have devoted a lot of time to it and dont see the solution, if anyone does thanks in aadvance.
if you see any other choice that is easy to prove this pls let me know because I dont have to do it by induction
 [/INST] Use the well-known inequality  proved here: https://math.stackexchange.com/a/1161287/148510.
For $0 < x < 1$,
$$1-x \leqslant -\ln x \leqslant \frac{1-x}{x}.$$
Then with $x = 1-1/k$, we have
$$\ln(1 -1/k) \geqslant \frac{-\frac1{k}}{1-\frac1{k}}=\frac{-1}{k-1}.$$
Hence,
$$(k-1)\ln(1 -1/k) \geqslant -1 \implies \left(1 - \frac1{k}\right)^{k-1}\geqslant e^{-1}.$$
For the other inequality, an application of the Bernoulli inequality yields
$$\left(\frac{k}{k-1}\right)^{k-1} = \left(1+\frac{1}{k-1}\right)^{k-1}\geqslant 1 + \frac{k-1}{k-1}= 2.$$
Hence,
$$\left(1 - \frac1{k}\right)^{k-1} =  \left(\frac{k-1}{k}\right)^{k-1} = \left(\frac{k}{k-1}\right)^{-(k-1)} \leqslant \frac1{2}.$$
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