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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

Yuriy S
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user1342645
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2 Answers2

1

Your last step is already very close to the solution:

$$\left( \frac{k}{1+k} \right)^k=\left( \frac{1}{1+\frac{1}{k}} \right)^k= \frac{1}{\left(1+\frac{1}{k}\right)^k} $$

Use the very well known limit and bounds for the number $e$ (for $k>1$):

$$2<\left(1+\frac{1}{k}\right)^k<e<\left(1+\frac{1}{k}\right)^{k+1}<4$$

The values on the left and right are for $k=1$, you can prove by induction that the sequence on the left is increasing while the sequence on the right side is decreasing.

$$\frac{1}{2}>\frac{1}{\left(1+\frac{1}{k}\right)^k}>\frac{1}{e}$$

I used $>$ and $<$ everywhere, since the equality on the right side would only appear for $k \to \infty$

Yuriy S
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0

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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