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Show that for natural $n \ge 2$ the following does hold:

$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$

First solution: $$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$ $$\iff$$ $$\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}>1-\frac{1}{n}$$ $$\left(1-\frac{1}{n}\right)^{n-1}\left(1+\frac{1}{n}\right)^{n}>1$$

By Bernoulli inequality: $$\left(1-\frac{1}{n}\right)^{n-1}\left(1+\frac{1}{n}\right)^{n}\ge2\left(1-\frac{n-1}{n}\right)=\frac{2}{n}$$

Which is not useful.

Second solution:

$$\left(1-\frac{1}{n^{2}}\right)^{n}=\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}\ge0\cdot2=0$$

Which is not useful.

Third solution:

$$\left(1-\frac{1}{n^{2}}\right)^{n}=\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}\ge2\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n-1}=\left(1+\left(-\frac{1}{2}\right)\right)^{n-1}$$$$\ge1-\frac{n-1}{2}=\frac{3-n}{2}$$

Final solution:

$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$ $$\iff$$ $$\left(1-\frac{1}{n^{2}}\right)^{n-1}\left(1+\frac{1}{n}\right)>1$$ By Bernoulli inequality: $$\left(1-\frac{1}{n^{2}}\right)^{n-1}\left(1+\frac{1}{n}\right)>1-\frac{n-1}{n^{2}}=1-\frac{1}{n}+\frac{1}{n^{2}}\ge\frac{1}{2}+\frac{1}{n^{2}}$$

Which is true to claim that $$\frac{1}{2}+\frac{1}{n^{2}}>1-\frac{1}{n}$$ Since for natural $n \ge 2$ we have that $$\frac{1}{n}\left(\frac{1}{n}+1\right)>\frac{1}{2}$$

Angelo
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masaheb
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1 Answers1

3

Show that for natural $n \ge 2$ the following does hold: $$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$

By using the Bernoulli’s inequality

$(1+x)^{n}>1+nx$

for any natural $\;n\ge2\;$ and for any $\;x>-1\;,$

with $\;x=-\dfrac1{n^2}>-1\;,\;$

we get that

$\left(1-\dfrac{1}{n^{2}}\right)^{n}>1+n\left(-\dfrac{1}{n^2}\right)=1-\dfrac1n\;,$

for any natural $\;n\ge2\;.$

Angelo
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